If$A \leq_T B$ is given, can you reduce $\overline{A}$ to $B$ and vice-versa
You say:

This is only possible if $A \le_m B$ exists and $B=\overline{B}$

But that condition simplifies to &quot;false&quot; since it can never the case that  $B=\overline{B}$.
Anyway the answer to both cases is &quot;yes&quot;.
If $A \le_T B$ then there is a Turing machine $T$ with oracle $B$ that decides $A$.
By complementing $T$'s output we obtain a Turing machine with oracle $B$ that decides $\overline{A}$  showing that $\overline{A} \le_T B$.
Moreover, consider a Turing machine $T'$ with oracle $\overline{B}$ that simulates $T$ except for the following: whenever $T$ invokes its oracle for $B$ on some input $w$, $T'$ invokes its oracle for $\overline{B}$ on $w$ and then complements the result. Clearly $T'$ still decides $A$, thus showing that $A \le_T \overline{B}$.