(1) Your proof that $\liminf A_n\subseteq \limsup A_n$ is correct.
Exercise 1: Under what conditions does equality hold in the above inclusion, i.e., under what conditions is it true that $\liminf A_n=\limsup A_n$?
(2) Note that $x\in \liminf A_n$ if and only if there exists a positive integer $N$ such that $x\in A_n$ for all $n\geq N$ if and only if there exists a positive integer $N$ such that $x\in \bigcap_{k=N}^{\infty} A_k$ if and only if $x\in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$.
I will leave the remaining questions as easy exercises (with similar solutions).
Exercise 2: Let $\{A_n\}$ be a sequence of measurable subsets of a measurable space $(X,\mu)$. Assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. Let $A=\limsup A_n$. Prove that $\mu(A)=0$. (Hint: Use the fourth assertion in your question, namely, use the characterization of $\limsup A_n$ in your question.)
Exercise 3: Give an example (in the context of Exercise 2) where $\lim_{n\to\infty} \mu(A_n)=0$ but that $\mu(A)>0$. Do not assume that $\Sigma_{n=1}^{\infty} \mu(A_n)<\infty$. (Hint: let $\{A_n\}$ be an appropriate sequence of intervals in $[0,1]$, for example.)
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 $x \in \mathbb{R}$, $\limsup_n E_n = \mathbb{R}$.
Best Answer
A member of $$ \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n $$ is a member of at least one of the sets $$ \bigcap_{n\ge N} A_n, $$ meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc. That means it's a member of all except finitely many of the $A$.
A member of $$ \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n $$ is a member of all of the sets $$ \bigcup_{n\ge N} A_n, $$ so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc. That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later. That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.