Closure under reversal of regular languages: Proof using Automata
You are asking two questions. I will only answer the second one. Recall that $L^* = \bigcup_{n=0}^\infty L^n$. I will show that $(L^k)^R = (L^R)^k$, and then it easily follows that $(L^*)^R = (L^R)^*$.

To show that $(L^k)^R = (L^R)^k$, notice that:
$$
(L^k)^R = \{ (w_1\ldots w_k)^R : w_1,\ldots,w_k \in L \} = \\
\{w_k^R \ldots w_1^R : w_k,\ldots,w_1 \in L \} = \\ \{z_k \ldots z_1 : z_k,\ldots,z_1 \in L^R \} = (L^R)^k.
$$
If you want to be more formal, you can prove this by induction on $k$. The crucial observation is $(xy)^R = y^R x^R$, which you can also prove by induction if you want to be very formal.