Is $L_2:=${$$|$L(M)=\overline{A_TM}$} (un-)decidable?
For $L_1:$
It's not decidable.
Define $C=\{A_TM\}$.
Now, by Rice's theorem we have that the language $L=\{&lt;M&gt;|L(M)\in C\}$ is undecidable. (By the extended theorem it's not even semi-decidable.)
But notice $L_1=L$ and therefore $L_1$ is undecidable.

For $L_2:$
It's decidable, although your solution is not correct.
Notice that $\overline{A_TM}$ is not semi-decidable, and therefore there is no TM $M$ where $L(M)=\overline{A_TM}$.
Thus, $L_2=\{&lt;M&gt;|L(M)=\overline{A_TM}\}=\emptyset$ and thus a TM that always rejects will decide $L_2$.