I like to think about the set $\liminf_n A_n$ as the set $\{ x \in X \mid x \in A_n \mbox { for all but finitely many } n \}$, and $\limsup_n A_n$ as the set $\{ x \in X \mid x \in A_n \mbox {for infinitely many } n \}$.
This follows from your definitions quite clearly: $x \in \limsup_n A_n$ iff $\forall N \exists n \ge N: x \in A_n$, which says that for every index $N$ we can find a larger index $n$ such that $x \in A_n$, which holds precisely when there infinitely many of such indices. On the other hand, $x \in \liminf_n A_n$ iff $\exists_N \forall n \ge N: x \in A_n$, which says that there is some index $N$ from which we know that $x \in A_n$ for all larger ones, so $x$ only possibly misses the $A_n$ with $ n < N$, so $x$ is in all but finitely many of the $A_n$. Note also that this makes the inclusion $\liminf_n A_n \subseteq \limsup_n A_n$ obvious.
Then in (a), the set alternate between $[0,1)$ and $[1,2]$, so indeed no point is in all but finitely many of them, as the odd- and even-indexed sets are disjoint, and all points in $[0,1) \cup [1,2] = [0,2]$ are in infinitely many of them. So I get $\liminf_n A_n = \emptyset$ and $\limsup_n A_n = [0,2]$.
In (b), we again have sets such that $E_{2n} \cap E_{2n+1} = \emptyset$ for all $n$, so that no point can be in all but finitely many of them, so $\liminf_n E_n = \emptyset$, so agreed.
Now, if $x \le 0$, then for $n \ge |x|$, $n$ even, we have that $x \in E_{n}$, so those $x$ are in infinitely many $E_n$, and if $x > 0$ then for $n > \min(\frac{1}{x}, x)$, $n$ odd, we know that $x \in E_{n}$ (as then $\frac{1}{n} < x < n$), again showing that $x$ is in infinitely many $E_n$. As $x \le 0$ or $x > 0$ holds for all $x \in \mathbb{R}$, $\limsup_n E_n = \mathbb{R}$.
Hint for (a): both sequences are monotone and bounded.
Hint for (b): you can explicitely find $\bar b_n$ and $\underline b_n$. Your intuition is correct, btw.
Hint for (c): $\bar b_n\ge\underline b_n$ always.
Hint for (d): $\bar b_n\ge a_n\ge\underline b_n$ to prove in one direction and use the fact that a convergent sequence satisfies the Cauchy criterion to prove in another direction.
Anothe way to understand $\limsup$ and $\liminf$ is to take all convergent subsequences; then find $\sup$ and $\inf$ of the set of their limits. This approach, for example, instantly gives the answer to (b).
Best Answer
The set $\liminf_nA_n$ is the set of those real numbers which belong to $A_n$, for every large enough $n$. So, $\liminf_nA_n=[0,2)$. In fact:
On the other hand, the set $\limsup_nA_n$ is the set of those real numbers which belong to infinitely $A_n$'s. Therefore, it is equal to $(-1,\infty)$.