If a functor is exact, does it always have an adjoint? If so, is the adjoint also exact? For the first statement, if a functor is exact, can it admit both a left and right adjoint? (since it's both left and right exact). Under what conditions can these statements hold?
 A: No. The basic problem is that in order for a functor to be a left resp. right adjoint it must preserve all colimits resp. limits, rather than only preserving finite colimits resp. limits (but this isn't sufficient either, in general). 
For example, let $M$ be a module over a commutative ring $k$. Then
$$\text{Hom}_k(M, -) : \text{Mod}(k) \to \text{Mod}(k)$$
is always continuous, and in fact it always has a left adjoint, namely $M \otimes_k (-)$. It is exact iff $M$ is projective (by definition). And it has a right adjoint iff $M$ is finitely generated projective, in which case the right adjoint is given by $\text{Hom}_k(M^{\ast}, -)$. So, for example, if $M$ is an infinitely generated free module, then $\text{Hom}_k(M, -)$ is exact but lacks a right adjoint. Similarly,
$$M \otimes_k (-) : \text{Mod}(k) \to \text{Mod}(k)$$
is always cocontinuous, and in fact it always has a right adjoint, namely $\text{Hom}_k(M, -)$. It is exact iff $M$ is flat (by definition). And it has a left adjoint iff $M$ is finitely generated projective, in which case the left adjoint is given by $M^{\ast} \otimes_k (-)$. So again we can take $M$ to be an infinitely generated free module, but we can also take $M$ to be a finitely generated flat module that isn't projective, although these are a bit harder to come by.
The sorts of theorems you want are called adjoint functor theorems. 
