Are $A$ and $B$ necessarily be decidable if $(A∩\overline{B})∪(\overline{A}∩B)$ is decidable and $A$ & $B$ being exhaustive?
It is possible for $A$ &amp; $B$ to be mutually exclusive and yet be exhaustive. Now this can happen even when one of them is not decidable. So option A.


The analysis above is correct except the last conclusion "So option A". We cannot exclude the possibility that $A$ and $B$ are both decidable. In fact, it happens in the current situation because of the extra condition, as shown in the latter part of this answer.



Hints, try using the following facts. If you do not know them, prove them first.


For any two languages $X$ and $Y$, if $X$ is exhaustive and $Y$ is decidable, then $X\cap Y$ is exhaustive. (Moreover, $X\setminus Y$ is exhaustive, too. We do not need this fact, though.)
For any language $X$, $X$ is decidable if (and only if) both $X$ and $\overline X$ are exhaustive.




Here is how we can find both $A$ and $B$ are decidable.

Since $A$ is exhaustive and $(A\cap \overline B)\cup(\overline A\cap B)$ is decidable, $A\cap ((A\cap \overline B)\cup(\overline A\cap B))=A\cap\overline B$ is exhaustive. Since $A\cup B=\Sigma^*$, $A\cap\overline B=\overline B$. Since $B$ and $\overline B$ are exhaustive, $B$ is decidable. 

Symmetrically, $A$ is decidable.