If A and B are nonzero c.e. degrees and the intersection of A and B is computable, must A and B be a minimal pair? Furthermore, does the converse hold?
By minimal pair I mean a pair (A,B) such that if some language C turing reduces to A and C turing reduces to B, then C is computable.
 A: I'm assuming that $A$ and $B$ should be sets, not degrees; otherwise per my comment the question doesn't make much sense.

Set-theoretic operations don't generally tell you much about reducibility and degrees. For example, let $X$ and $Y$ be any two sets and consider $EX=\{2x: x\in X\}$ and $OY=\{2y+1: y\in Y\}$. Then $EX\equiv_TX, OY\equiv_TY$, and $EX\cap OY=\emptyset$. So the set-theoretic intersection tells us nothing.
Similarly, given any two sets $X$ and $Y$ the sets $\{2z: z\in X\}\cup\{2z+1: z\in\mathbb{N}\}$ and $\{2z+1: z\in Y\}\cup\{2z: z\in\mathbb{N}\}$ are Turing equivalent to $X$ and $Y$ respectively but their union is everything, so the union tells us nothing.
In particular, the answer to both questions you ask is no.

The situation is somewhat more interesting with the symmetric difference. Here things are more subtle: sometimes symmetric difference is well-defined on degrees. A bit about this is said here; Uri Andrews, Peter Gerdes, Joe Miller, Mariya Soskova, and I worked a bit more on the topic, but that hasn't been written up yet.
