First, you might also want to take a look at this answer to a similar question.
Okay: the first description assumes that there is some sort of notion of "accumulation point" at work in the set $X$, as you surmise; this may be derived from a topology.
The second description talks about limit points, but you can apply it to any set by endowing the set with the discrete topology (every subset is open, every subset is closed). If you do that, then the definition is the usual definition of limit superior of a sequence of sets: it is the collection of all points that are in infinitely many of the terms of the sequence, while the limit inferior is the collection of all points that are in all sufficiently large terms of the sequence.
The "second way" of defining it is in terms of unions and intersection. If $\{X_n\}_{n\in\mathbb{N}}$ is a family of sets, then
\begin{align*}
\limsup_{n\in\mathbb{N}} X_n &= \bigcap_{n=1}^{\infty}\left(\bigcup_{j=n}^{\infty} X_j\right)\\\
\liminf_{n\in\mathbb{N}} X_n &= \bigcup_{n=1}^{\infty}\left(\bigcap_{j=n}^{\infty} X_j\right).
\end{align*}
This coincides with the notion of the limit superior being the set of all limit points of infinitely many terms in the sequence, under the discrete topology; and the limit inferior being the set of all limit points of all sufficiently large-indexed terms of the sequence (again, under the discrete topology).
The notion of "accumulation point" in the first description is more informal. If you are working with a topological space, then it is limit points as described above and by "accumulation set" you should read "set of all limit points".
For your third point, in order to be able to talk about joins and meets you need to have some sort of complete lattice order on your set, so that you can talk about those infinite meets and infinite joins; this is the case, for instance, in the real numbers; appropriately interpreted, you do get essentially the definition you propose, though you need to tweak it a bit in order to actually get what the actual definition is (see the other answer quoted above); you don't actually work with the points themselves, but with a slightly different set determined by the points.
I think that the previous answer linked to answers essentially your fourth point, of how to interpret limit superior and limit inferior of a sequence of points as a special case of limit superior and limit inferior of sets; but if this is not the case, point it out and I'll try to answer it de nuovo.
Use the definition(s) of $\limsup$ and $\liminf$. For example, the limit superior of a sequence, is defined as $\limsup a_n = \sup_{k \geq 1} \inf_{n \geq k} a_n$, and for $\liminf$ the $\sup$ and $\inf$ are switched.
Here is a better way of understanding these concepts. If $a_n$ is any sequence, then $b_n = \sup_{k \geq n} a_k$(in the extended reals) is an decreasing sequence(because the set $\{k \geq n\}$ keeps shrinking as $n$ grows larger). Therefore, every decreasing sequence either has a limit(if it is bounded) or goes to $-\infty$. The limit superior is defined as the limit of the decreasing sequence, if it exists, else it is $\pm\infty$ depending upon where the sequence goes (Note : $\limsup$ is defined over the extended reals, whence it can possibly take the values $\pm \infty$).
Similarly, the limit inferior would be defined as follows : $c_n = \inf_{k \geq n} a_k$ is an increasing sequence. So either it has a limit, which we call the limit inferior, or it goes to $\pm\infty$, whence the limit inferior would also be $\pm \infty$.
With this outlook, let us look at the questions we are given.
What is $b_n$ here? $b_n = \sup_{k \geq n} a_k$. But then $a_k$ is an increasing sequence that increases to $\infty$. So, $b_n$ is $\infty$, because the supremum is greater than each element, but the elements are themselves arbitrarily large. In particular, the limit superior is also $+\infty$. Following a similar logic gives you $\liminf a_n = -\infty$.
Of course, $a_n$ is constant, so $b_n = \sup_{k \geq n} a_k = 2$ (the supremum of the set $\{2\}$ is just $2$!) and the limit of the constant sequence $2$ is $2$. So the limit superior (analogously, limit inferior) are both equal to $2$.
- $n \sin (\frac{n \pi}{2})$.
This is a weird sequence. Look at its terms : $1,0,-3,0,5,0,-7,0,...$. In other words, this sequence can be rewritten like this : $$a_n = \begin{cases}
0 \quad n \equiv 0(2) \\
n \quad n \equiv 1(4) \\
-n \quad n \equiv 3(4)
\end{cases}$$
where $n \equiv m(k)$ means that $n-m$ is a multiple of $k$. Now, what is $b_n = \sup_{k \geq n} a_k$? If you look at the set $\{a_k,a_{k+1},a_{k+2},...\}$ then this contains a strictly increasing sequence of positive numbers, that comes from the $n$ part of the definition above. Therefore, $b_n$ is actually infinite, and so there, is the limit superior. Similarly, the $-n$ part contributes to the limit inferior becoming $-\infty$.
- $0,1,0,1,2,0,1,3,0,1,4...$
This is the trickiest of the lot. Look at $b_n = \sup_{k \geq n} a_k$. The sequence $\{a_k,a_{k+1},a_{k+2},...\}$ contains a strictly increasing sequence of positive numbers. Hence, its limit superior is $+\infty$.
On the other hand, the limit inferior? Observe the sequence $c_n = \inf_{k \geq n} a_k$. Well, observing the sequence $\{a_k,a_{k+1},...\}$, we see that $0$ must be there in this set. But then, $0$ is also the smallest element of this set! So, $c_n = 0$ for all $n$, so the limit inferior is $0$.
EDIT : I'll give you a bonus here.
The sequence $-1,-2,-3,...$ will have both $\liminf$ and $\limsup$ as $-\infty$. Try to show this from the definitions. This is a quirky point, and a nice example.
EDIT AGAIN : The new definition : the limit inferior of $x_n$ is the infimum of the set $\{v\}$ such that $v < x_n$ for at most finitely many $x_n$.
This is exactly the set $V$ of all numbers which are greater than or equal to all the numbers that appear "after some time" in the sequence i.e. there is $N$ such that for all $n > N$, $v > x_n$.
For example, lets see this for the last example. In $0,1,0,1,2,0,1,3,0,1,4,...$, what is the set $V$? Which is that element, which is greater than or equal to all elements "after some time"? I claim, that there is none of them. This is because, this sequence has a subsequence that is exactly the natural numbers, but no fixed number can be greater than all the natural numbers. So here, the limit superior is $+\infty$.
Similarly, the limit inferior is defined as the supremum over all $\{w\}$ such that $w > x_n$ for only finitely many $n$. So, what is that set here?
Well, W definitely contains all the negative integers and $0$. But anything more? No, because there are infinitely many zeros in the sequence, so anything $ a > 0$ will be greater than all those terms of the sequence which are $0$, but there are infinitely many of them so $a \notin W$! The supremum of $(-\infty,0]$ is $0$, so there you have it.
Best Answer
I suspect that if you have a complete lattice $X$ and you endow it with the order topology (where a subbasis of the topology is given by the sets of the form $(a,\infty) = \{x\in X\mid a\lt x\}$ and $(-\infty,b) = \{x\in X\mid x\lt b\}$ with $a,b\in X$ arbitrary), then 1, 2(1), and 2(2) will be equivalent in the sense that one exists if and only if the others exists and in that case they will be equal (the only caveat is that we have to be careful with 2(1) in the case of unbounded sequences; technically, for instance, if you take $1,2,1,3,1,4,1,5,1,6,\ldots$, the only converging subsequences converge to $1$, but the sups are never defined, so you could interpret 2(1) as saying that the limit superior is $1$, and 2(2) as saying it does not exists).
However, I am having trouble proving it in full generality; the equivalence of 2(1) and 2(2) is not difficult if you assume the set is totally ordered and complete, so that suprema exist for any bounded set (throw in an über-sup and an über-inf like $\infty$ and $-\infty$ to the reals if necessary). I'm somewhat unhappy about this, but I figure something is better than nothing here. I've been toying with this and going down a fair number of silly alleys along the way, so maybe after it is written down I'll see that it works in any lattice; however, right now I am using the fact that any sequence contains a monotone subsequence, and this requires total order, not just lattice order. (I'm also using the fact that the order topology is Hausdorff for totally ordered sets; I don't know if that works for lattices).
Note that a totally ordered set that has the order topology is necessarily Hausdorff: given $x\neq y$, say $x\lt y$. If there exists $z$ such that $x\lt z\lt y$, then the sets $(-\infty,z)$ and $(z,\infty)$ provide disjoint open neighborhoods of $x$ and $y$, respectively; if there is no element of $X$ strictly between $x$ and $y$, then $(-\infty,y)$ and $(x,\infty)$ provide the necessary disjoint neighborhoods.
Theorem. Suppose $X$ is a totally ordered set such that every nonempty subset has a supremum. If we endow $X$ with the order topology, then definitions 2(1) and 2(2) are equivalent.
Proof. Say $\lim\limits_{n\to\infty}\sup\{x_m\mid m\geq n\}$ is equal to $L$. This means that for every open neighborhood $U$ of $L$, there exists a $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$, and in particular for every $k\geq K$ there exists $m\geq k$ such that $x_m\in U$.
First, I claim that the set of limit points of subsequences of $x_n$ is bounded above by $L$. Indeed, suppose that you have a subsequence that converges to $M$, with $M\gt L$. Let $U$ and $V$ be neighborhoods of $L$ and $M$ that are disjoint; intersect $U$ with $(-\infty,M)$ and $V$ with $(L,\infty)$, if necessary. Then there exists $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$; in particular, for all $k\geq K$, we have $x_k \leq\sup\{x_m\mid m\geq k\}\notin V$; therefore, since the sequence is eventually not in $V$, it is not infinitely often in $V$, and therefore there can be no subsequence that converges to $M$. This contradicts the choice of $M$, so we conclude that $M\leq L$, as claimed.
Since every sequence contains a monotone subsequence, and monotone subsequences in complete sets have limits (their sup if they are increasing, their inf if they are decreasing), the set of limit points of subsequences is nonempty; therefore, since it is bounded above by $L$, we know that $$\sup\{\text{limit of any subsequence of }\{x_n\}\text{ that converges}\} \leq L=\lim_{n\to\infty}\sup\{x_m\mid m\geq n\}.$$
Suppose that the supremum of the limits is $M$, and is strictly smaller than $L$; let $V$ be an open neighborhood of $M$ contained in $(-\infty,L)$, $U$ an open neighborhood of $L$ contained in $(M,\infty)$, with $U\cap V=\emptyset$. We know that there exists $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$; in particular, for all $k\geq K$ there exists $m\geq k$ such that $x_m\in U$. Thus, we can find a subsequence of $\{x_n\}$ that is contained in $U$; this subsequence contains a monotone subsequence, which therefore coverges to some point in the closure of $V$; the closure of $V$ is contained in the complement of $U$ and of $(-\infty,M)$, and therefore the closure of $V$ does not include $M$ and consists only of points strictly larger than $M$; that is, the limit of this monotone subsequence is strictly larger than $M$, contradicting the choice of $M$. Thus, $M\geq L$, giving equality. QED
I don't know about 3 and 4, I'll think about 5, though again I expect you'll need the topology to be related to the order.
Added.
In any lattice, with definition 1, you always have $\liminf x_n\leq \limsup x_n$. To see this, note that for every $k,m$, you have $$\mathop{\wedge}\limits_{n\geq k} x_n \leq \mathop{\vee}\limits_{n\geq m}x_n$$ because if $r\geq \max\{k,m\}$, then you have $$\mathop{\wedge}\limits_{n\geq k}x_n \leq x_r \leq \mathop{\vee}\limits_{n\geq m}x_n.$$ Therefore, each $\mathop{\vee}\limits_{n\geq m}x_n$ is an upper bound for all of $\mathop{\wedge}\limits_{n\geq k}x_n$, hence for each $m$ we have $$\liminf x_n = \sup\mathop{\wedge}\limits_{n\geq k}x_n \leq \mathop{\vee}\limits_{n\geq m}x_n.$$ Therefore, $\liminf x_n$ is a lower bound for each of the joins, hence is less than or equal to their inf; that is, $$\liminf x_n \leq \inf\mathop{\vee}\limits_{n\geq m}x_n = \limsup x_n.$$ (The same is also true for definition 2(1), for silly reasons: the limit superior is the supremum of a particular nonempty set, the limit inferior is the infimum of that same set, and since the set is nonempty, then the infimum is less than or equal to the maximum).
Let $L=\liminf x_n$ and $M=\limsup x_n$.
Suppose that $\liminf x_n \lt \limsup x_n$, and that you endow $X$ with the order topology inherited from the complete lattice structure. Because $\liminf x_n$ and $\limsup x_n$ are comparable, then we can prove the existence of disjoint open neighborhoods that separate them the same way I did for a totally ordered set: if there is an element $z\in X$ such that $L\lt z\lt M$, then the open sets $(-\infty,z)$ and $(z,\infty)$ are disjoint open neighborhoods of $L$ and $M$, respectively; and if there is no such element, then $(-\infty,M)$ and $(L,\infty)$ are disjoint open neighborhoods that separate them. Taking disjoint neighborhoods of $L$ and $M$ you can obtain subsequences that are eventually not in the corresponding neighborhood, so if $L\lt M$ then there are subsequences that do not converge.
If $L=M$, on the other hand, then let $U$ be an open neighborhood of $L=M$ that is an interval; because the sequence $w_k = \wedge\{ x_m\mid m\geq k\}$ is nondecreasing, and $U$ is an interval containing $L$, there exists $K$ such that for all $k\geq K$ you have $w_k\in U$. Likewise, by considering the nonincreasing sequence $z_k\vee\{x_m\mid m\geq k\}$ which converges to $M$, there exists $H$ such that for all $h\geq H$ we have $z_h\in U$. Therefore, for every sufficiently large $k$ we have that both the meet and the join of $\{x_m\mid m\geq k\}$ lie in the interval $U$, then all the terms lie in the interval $U$; that is, the sequence is eventually in $U$. That is, given any open neighborhood of $L=M$, then the sequence is eventually in $U$; this means that it converges to $L=M$; and from here it is easy to verify that every subsequence also converges to $L$.
Therefore, $\limsup x_n = \liminf x_n$ if and only if there exists $L$ such that every subsequence converges to $L$, if and only if there exists $L$ such that the sequence converges (in the topological sense) to $L$.
I don't know if the limit needs to be unique (in a topology, you can have a sequence converge to two different things if the topology does not separate the points enough).
For 2(1), assuming the topology is the order topology, and $X$ is totally ordered, if $L\lt M$, then we can find intervals $(-\infty,a)$ and $(b,\infty)$ that are disjoint, $L\lt a$ and $b\lt M$, as above. Then there is a subsequence that converges to an element of $(-\infty,a)$ (in fact, of $[L,a)$) and a subsequence that converges to an element of $(b,\infty)$ (in fact, of $(b,M]$). Therefore, the original sequence is infinitely often in $(-\infty,a)$, and infinitely often in $(b,\infty)$; since these two are disjoint, the sequence cannot converge (if it converges to $x$, then for every open neighborhood of $x$ the sequence is eventually in the neighborhood; but separate $x$ from either $L$ or $M$ and get that the sequence is eventually not in the corresponding neighborhood of $x$). Thus, if $L\lt M$ then the sequence does not converge.
If $L=M$, then every converging subsequence converges to $L$; but if the set is not complete, then you'll run into trouble: take for example $X=\mathbb{Q}$, a sequence of rationals that converges to $\sqrt{2}$, and interweave it with the constant sequence $1$. If you take a subsequence, then either it contains infinitely many terms of the sequence that converges to $\sqrt{2}$, and therefore does not converge in $X$, or else it contains only finitely many terms of the sequence, and therefore converges to $1$; so the set of limits of converging subsequences is $\{1\}$, hence the limits inferior and superior are equal under definition (1), but the sequence does not converge; if the space $X$ is complete relative to the order topology (so that every set has a supremum and every set has an infimum), then we have the equivalence between definition 2(1) and 2(2) by the above, and then we are essentially in the situation of 1 that I just dealt with to show that the sequence converges to $L=M$ in the sense of 2(2), and hence in the sense of 2(1).