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
Let $\{a_n\}$ be a bounded sequence. Then we define the sequences $\{a_n^+\}$ and $\{a_n^-\}$ by
$$a_n^+=\sup\{a_n,a_{n+1}\dots\}$$
$$a_n^-=\inf\{a_n,a_{n+1}\dots\}$$
We (may) then define
$$\lim a_n^+=\limsup a_n$$ $$\lim a_n^-=\liminf a_n$$
Now, you need two things to work this out:
$(1)$ Let $A$ be any bounded nonempty subset of $\Bbb R$. Then
$$\inf A\leq \sup A$$
$(2)$ Let $\{\alpha_n\}$ be a sequence such that $a_n\geq0 $ for each $n\in \Bbb N$. Then $$\lim a_n\geq 0$$
With $(1)$ you should show $$a_n^-\leq a_n^+$$ for each $n\in \Bbb N$. Monotone convergence says both $\{a_n^+\}$ and $\{a_n^-\}$ converge, since they are bounded (above/below) and are monotone (increasing/decreasing)$^{(*)}$. But $$a_n^+- a_n^-\geq 0$$
for each $n\in \Bbb N$, so use $(2)$ to show
$$\lim a_n^+-\lim a_n^-\geq 0$$
that is:
$$\liminf a_n\leq \limsup a_n$$
$(*)$ To prove this, you need to show that if $A\subseteq B$, then $$\sup A\leq \sup B$$ $$\inf A\geq \inf B$$ Then, observe that
$$\{a_{n+1},a_{n+2},\cdots\}\subseteq \{a_n,a_{n+1},a_{n+2},\cdots\}$$