Regular Closure

Given any two regular languages $L_1$ and $L_2$ , their intersection, $L_1\cap L_2$ , is also regular. From a simple inductive argument, it follows that $[1]$ is true.

$[3]$ is not necessarily true. Remember that any language that contains exactly one string is regular. Set $L_i=\{0^i1^i\}$ (the set containing the string that has $i$ zeros followed by $i$ ones). $\displaystyle\bigcup_{i=1}^\infty L_i$ is therefore $\{0^k1^k| k\geq 1\}$ . Call this language $S$ . $S$ is a non-regular language as shown in the wiki linked in the problem.

$[2]$ is also not true. Complement all of the $L_i$ s in the previous paragraph. Since regular languages are closed under complementation, we are left with another sequence of regular languages the infinite intersection of which gives us the set $\overline{S}$ . Now, $\overline{S}$ can't be regular since its complement, $S$ is non-regular.