Intuition about decidability
For the first undecidable language $L_1$ you can apply the Rice's theorem:

Theorem (Rice’s Theorem): Let L be a language of the form

$L = \{ \langle M \rangle | L(M)\mbox{ has some property }P\}$,where


P is non-trivial, i.e. there exist at least one machine M such that $\langle M \rangle \in L$, and at least one machine $M$ such that $\langle M \rangle \notin L$.
P is indeed a property of the language of TMs, i.e. whenever $L(M_1) = L(M_2)$,
we have $\langle M_1 \rangle \in L$ if and only if $\langle M_2 \rangle \in L$.


Then L is undecidable.

You can pick a Turing machine that decides $A_0$ and one that decides $\overline{A_0}$, so $L_1$ asserts a nontrivial property of $\langle M \rangle$, and so it is undecidable.

For the second decidable language $L_2$ the machine $M_0$ is fixed (it is not part of the language).
So $L(M_0)$ is regular or it is not regular. If it is not regular then $L_2 = \emptyset$ (trivially decidable). If $L(M_0)$ is regular then there is a DFA $A_{M_0}$ such that $A_{M_0} = L(M_0)$ and the language $L_2$ becomes:

$L_2 = \{ \langle A \rangle | L(A) = L(A_{M_0}) \}$

but the equivalence of two DFAs is decidable, so $L_2$ is decidable.