Is the power of a regular language regular? Is the root of a regular language regular?
The language $L_2$ is not necessarily regular. Indeed, consider $A = a^*b$. If $L_2$ were regular, then so would the following language be:
$$L_2 \cap a^*ba^*b = \{ a^nba^nb : n \geq 0 \}.$$
However, this language is not regular (exercise).

In contrast, the language $L_1$ is regular. We can see this by constructing a DFA for it. Let the DFA for $L_1$ have states $Q$, initial state $q_0$, accepting states $F$, and transition function $\delta$. The states of the new DFA are labeled by functions $Q \to Q$. The idea is that the new DFA is at state $f\colon Q \to Q$ if the original DFA moves from state $q \in Q$ to state $f(q)$ after reading $w$ (i.e., if $\delta(q,w) = f(q)$ for all $q \in Q$). The initial state is the identity function. When at state $f$ and reading $\sigma$, we move to the state $g$ given by $g(q) = \delta(f(q),\sigma)$. A state is accepting if $f^{(n)}(q_0) \in F$ for some $n \geq 0$.