When does tensor product have a (exact) left adjoint?

For question 2, $F\otimes_A -$ has a left adjoint iff $F$ is finitely generated, and the left adjoint is always exact. For if $(f_i)\in F^I$ is an element of an infinite product of copies of $F$, then it is easy to see $(f_i)$ is in the image of the canonical map $F\otimes_A A^I\to F^I$ iff $\{f_i\}$ is contained in a finitely generated submodule of $F$. If $F\otimes_A -$ has a left adjoint, then it this canonical map must be an isomorphism, and it follows that $F$ must be finitely generated. Conversely, if $F$ is finitely generated (and hence projective), $F^{\vee}\otimes_A -$ is adjoint to $F\otimes_A -$ on both sides.

(By a similar argument using instead the injectivity of the map $F\otimes_A A^I\to F^I$, you can show that $F$ must be finitely presented, even if you don't assume $A$ is Noetherian. So for non-Noetherian $A$, you still get that $F\otimes_A -$ has a left adjoint iff $F$ is finitely presented and flat, or equivalently finitely generated and projective.)

Note that in this case you can also easily see directly that $F\otimes_A -$ preserves injectives, since $F$ is a direct summand of a finitely generated free module and tensoring with a finitely generated free module obviously preserves injectives. In general, however, $F\otimes_A -$ might preserve injectives without having a left adjoint. For instance, it is a well-known theorem that a ring is Noetherian iff any (possibly infinite) direct sum of injective modules is injective. So since you're assuming $A$ is Noetherian, tensoring with any free module preserves injectives, and hence so does tensoring with any projective module.

Let me remove most of your assumptions and work with an arbitrary module $M$ over an arbitrary commutative ring $R$. We'd like to know when the functor $M \otimes_R (-)$ has a left adjoint.

The answer is iff $M$ is finitely presented projective, in which case the left adjoint is $M^{\ast} \otimes_R (-)$ where $M^{\ast} = \text{Hom}_R(M, R)$. You can extract a proof from this blog post. Since $M^{\ast}$ is also finitely presented projective, this functor is always exact.