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My attempt at finding $$a.) \underline{\lim_{n \to \infty}}A_n$$ and $$b.) \overline{\lim_{n \to \infty}}A_n$$ $A_n=B$ for $n$ even; $A_n=C$ for $n$ odd.
$a.)=\bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty}A_n=\emptyset$
$b.)=\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty}A_n=\{B,C\}$
It’s helpful to remember that $\limsup_nA_n$ is the set of points that are in infinitely many of the sets $A_n$, and that $\liminf_nA_n$ is the set of points that are in all but at most finitely many of the sets $A_n$. This shows right away that your answer $\{B,C\}$ cannot be right: it’s not a set of things that are elements of the sets $A_n$ at all!
Suppose that $x\in B$; then $x\in A_{2n}$ for every $n$, so $x$ is in infinitely many of the sets $A_n$, and therefore $x\in\limsup_nA_n$. Similarly, suppose that $x\in C$; then $x\in A_{2n+1}$ for every $n$, so $x$ is in infinitely many of the sets $A_n$, and once again we find that $x\in\limsup_nA_n$. This shows that $B\cup C\subseteq\limsup_nA_n$. On the other hand, if $x\notin B\cup C$, then $x$ isn’t in any of the sets $A_n$, so it certainly isn’t in infinitely many of them, and therefore $x\notin\limsup_nA_n$. This shows that $\limsup_nA_n\subseteq B\cup C$ and hence that $\limsup_nA_n=B\cup C$.
Now what points are in all but finitely many of the sets $A_n$? If $x$ is in all but finitely many of the sets $A_n$, then $x$ must be in every $A_n$ from some point on, and therefore $x$ must be in both $B$ and $C$. In other words, $\liminf_nA_n\subseteq B\cap C$. On the other hand, it’s clear that if $x\in B\cap C$, then $x\in A_n$ for every $n$, so $x\in\liminf_nA_n$. Thus, $\liminf_nA_n=B\cap C$.
If you really want to do so, you can turn these arguments into technical calculations using the $\bigcup\bigcap$ and $\bigcap\bigcup$ definitions of $\liminf$ and $\limsup$, but I really do think that the ideas are easier to work with first in this form.
