Check the Regularity of the folowing Languages
Your reasoning regarding $L_1$ is wrong. What you show is that for each $n$, the following language is regular whenever $A$ is regular:
$$
L_1^{(n)} = \{ x : x^n \in A \}.
$$
The language $L_1$ is the infinite union of $L_1^{(n)}$ for all $n \geq 0$. Unfortunately, regular languages are not closed under infinite union (for example, every unary language is an infinite union of singleton languages).

Nevertheless, $L_1$ is regular. Given a DFA $\langle Q,q_0,F,\delta \rangle$ for $A$, we construct a DFA for $L_1$ whose set of states is the set of all functions $Q \to Q$. The initial state $id$ is the identity function, and we construction the automaton so that $\delta_{L_1}(id,w)$ is the function that maps $\sigma \in Q$ to $\delta(\sigma,w)$. This gives us enough information to determine whether any power of $w$ belongs to $A$.

The language $L_2$, in contrast, is not necessarily regular. For example, if $A = ab^*$ then $L_2 \cap ab^*ab^* = \{ ab^n ab^n : n \geq 0 \}$, which isn't regular.