Determining if given languages are regular or recursively enumerable
Your intuition is entirely correct; this solution is nonsense. $L_1$ isn't a regular language; it's not even a decidable language, by Rice's Theorem, and also not recognizable (aka recursively enumerable).

$L_2$ is in fact recognizable. The following algorithm recognizes it:


  Take an encoded Turing machine $M$ as input. First simulate it on every input of length ≤ 1 for 1 step, then on every input of length ≤ 2 for 2 steps, and so on. Keep track of how many distinct inputs it's ever accepted. If this number ever exceeds 2016, return True.


But as it turns out, none of this matters to the actual problem. Note that (C) is taking the union of a language with its complement, and the union of any language with its complement is $\Sigma^*$. This language is regular, thus decidable, thus recognizable. So the correct answer is (C).

EDIT: Of course, the intersection of a language with its complement is $\varnothing$, which is also regular, thus decidable, thus recognizable. So (B) is also true.