Why is $A \leq_T \bar{A}$? I know there's some easy proof for this, but why is $A \leq_T \bar{A}$ ? ... Suppose that $\bar{A}$ is c.e. and could loop, does this mean that the oracle $TM$ $M^{\bar{A}}$ could also loop? 

But shouldn't $\leq_T$ mean that $M^{\bar{A}}$ be decidable?
A language $A$ Turing-reduces to $\overline{A}$ since a Turing machine with an $\overline{A}$-oracle can compute $A$. On input $x$, it determines whether $x \in \overline{A}$ using the oracle, and then outputs the reverse answer.