The four "approximate derivatives" are usually called Dini derivatives. They encapsulate best approximations to the usual derivative (and are always defined even if the latter fails to exist) in four ways: they are upper and lower approximations to the derivative from the left and the right.
If you know what the graph of $x\sin{\frac{1}{x}}$ looks like then it should be rather easy to imagine what they are supposed to be:
Note that the graph of $x \sin{\frac{1}{x}}$ touches each of the four rays issuing from the origin infinitely often. Every ray with smaller slope will intersect infinitely often, while every ray with larger slope will be too far away
Concerning the definition of $\limsup$, remember the reason for the notation. If $(x_n)_{n \in \mathbb{N}}$ is a sequence of real numbers, then $\limsup_{n\to \infty} x_{n} = \lim_{n \to \infty} \sup_{k \geq n} x_{k}$. If the sequence is bounded then this limit exists because $s_{n} = \sup_{k \geq n} x_{k}$ is a decreasing sequence and because the real numbers are complete. You can also write $$\limsup_{n \to \infty} x_{n} = \inf_{n \geq 0} \sup_{k \geq n} x_{k}.$$ As for the generalization to functions remember that a sequence is a function $x(n)$ from the natural numbers to the real numbers. Now if $f: (0,\varepsilon) \to \mathbb{R}$ is a function you can generalize the last expression for the $\limsup$ by writing
$$\limsup_{h \searrow 0} f(h) = \inf_{h \geq 0}{\sup_{x \in (0, h)}} f(x).$$
If you understand the definition of infimum and supremum properly, you should have no problem in solving the following:
Exercise: If $a = \limsup_{h \searrow 0} f(h)$ then there exists a sequence $x_{n} \searrow 0$ such that $a = \limsup_{n \to \infty} f(x_{n})$ and for every other sequence $\limsup_{n \to \infty} y_{n} \leq a$.
In other words, in order to calculate $\limsup_{h \searrow 0} f(h)$, find the maximum possible $\limsup_{n \to \infty} f(x_{n})$ among all sequences $x_{n} \searrow 0$.
You ask whether or not your reasoning is correct. It is close, but not quite there.
Let's start from the beginning. We have $a_n = (-1)^n + 1/n$. Now, we can define (as you did), the sequence of sequences $$A_n = \{a_k\colon k \geq n\} = \{a_n, a_{n+1}, \ldots\}.$$
Note that each $A_n$ is not a number, but rather a sequence of numbers. (This makes your guess for $A_n$ incorrect.)
Now, by definition
$$b = \limsup a_n = \inf_{n\geq 1}\ \sup_{k\geq n} a_k = \inf_{n\geq 1}\sup\{a_k\colon k \geq n\} = \inf_{n\geq 1}(\sup A_n) = \inf_{n\geq 1}\, b_n$$
$$c = \liminf a_n = \sup_{n\geq 1}\ \inf_{k\geq n} a_k = \sup_{n\geq 1}\inf\{a_k\colon k \geq n\} = \sup_{n\geq 1}(\inf A_n) = \sup_{n\geq 1} \,c_n,$$
where we let $b_n = \sup A_n$ and $c_n = \inf A_n$, as you did.
So let's compute $b_n = \sup A_n$ and $c_n = \inf A_n$.
Claim: We claim that $$b_n = \sup A_n = \begin{cases} 1 + \frac{1}{n} \ \ \ \text{ if } n \text{ is even} \\ 1 + \frac{1}{n+1} \ \ \ \text{ if } n \text{ is odd} \end{cases}$$
and
$$c_n = \inf A_n = -1 \ \ \text{ for all } n \geq 1.$$
Proof: Let's look at $b_n$ first. If $n$ is even, then $$A_n = \left\{1 + \frac{1}{n}, \ -1 + \frac{1}{n+1}, \ 1 +\frac{1}{n+2}, \ldots\right\}.$$
It is clear that $A_n$ contains a largest element, namely $1 + \frac{1}{n}$. (If you don't find this fully precise, I invite you to fill in the details.) Therefore, $\sup A_n = 1 + \frac{1}{n}$ when $n$ is even.
If $n$ is odd, then $$A_n = \left\{-1 + \frac{1}{n}, \ 1 +\frac{1}{n+1}, \ -1 + \frac{1}{n+2}, \ldots\right\}.$$
Again, $A_n$ contains a largest element, namely $1 + \frac{1}{n+1}$, so that $\sup A_n = 1 + \frac{1}{n+1}$. This verifies the claim about $b_n$.
Let's look at $c_n$ now. We want to show that $\inf A_n = -1$, i.e. that $-1$ is the greatest lower bound of $A_n = \{a_k\colon k\geq n\}$.
Now, since $-1 < a_k$ for every $k \geq n$ (you can check this), it follows that $-1$ is a lower bound of $A_n$. To show that it is the greatest lower bound, we need to show that if $q > -1$, then $q$ is not a lower bound for $A_n$.
Let $q > -1$. Choose $k \geq n$ large enough so that $k > \frac{1}{q+1}$. Then $k(q+1) > 1$, so $q+1 > \frac{1}{k}$, so $q > -1 + \frac{1}{k}$. But $-1 + \frac{1}{k}$ is an element of $A_n$ (since $k \geq n$), and we just showed that it is less than $q$. Therefore, $q$ is not a lower bound for $A_n$. We therefore conclude that $\inf A_n = -1$ (for all $n \geq 1$).
This proves the claim.
Finally, it follows from our claim (I leave the details to you) that $$b = \limsup a_n = \inf_{n\geq 1}\ b_n = 1$$ and $$c = \liminf a_n = \sup_{n\geq 1} \ c_n = -1.$$
Best Answer
$$D^+f(0)=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{f(t)-f(0)}{t}=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{t\sin \frac1t}{t}=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\sin\frac{1}{t}$$ Now no matter how small $h$ gets, there always exist a point $t$ in $(0,h]$ so that $\sin\frac{1}{t}=+1$ (take $t=\frac{1}{n\pi +\frac{\pi}{2}}$ for sufficiently large $n$) Therefore, $D^+f(0)=+1$. Similarly $D^{-}f(0)=-1$.
At all points $x\neq 0$, $x\sin\frac{1}{x}$ is differentiable and as thus $$D^+f(x)=D^-f(x)=f^{\prime}(x)=\sin \frac{1}{x}-\frac{1}{x}\cos\frac1x$$
For $g$: If $x\in \mathbb{Q}$,$$D^+g(x)=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{g(x+t)-g(x)}{t}=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{\chi_{\mathbb{Q}}(x+t)-1}{t}$$ Now $$\frac{\chi_{\mathbb{Q}}(x+t)-1}{t}=\begin{cases} 0&\mbox{if } t\in \mathbb{Q}\\ \frac{-1}{t}&\mbox{if } t\notin \mathbb{Q}\end{cases}$$ The supremum is achieved when $t<0$ and $t\notin \mathbb{Q}$ which means $0<-t\le h\Rightarrow -h\le t<0$. No matter how small $h$ gets there always exists a point $t\in [-h,0)\cap \mathbb{Q}^c$ (density argument) and so $$D^+g(x)=\lim_{h \to 0}\sup\limits_{0<|t|\leq h}\frac{\chi_{\mathbb{Q}}(x+t)-1}{t}=\lim_{h \to 0}\sup\limits_{-h\le t<0}\frac{-1}{t}=+\infty$$ All other cases are done in the same spirit