Does the adjoint relationship between Tensor and Hom functors give an adjoint relationship between Ext and Tor? The question is as in the title; I don't really have a more specific question, I am more interested in a "yes, here is an example of how it is useful," a "yes, but it really isn't useful," or a "no, here is why."
The context is we are covering homological algebra out of Lang for our algebra course. The problem is I still wasn't completely comfortable with the categorical properties of the Hom and Tensor functors before moving on to Ext and Tor, so I am trying to make some connections between the two notions to try and solidify my understanding. 
I am comfortable with the adjoint relationship between Hom and Tensor; namely, $$Hom(Y\otimes X, Z)\cong Hom(Y, Hom(X,Z))$$
So potentially a candidate would be 
$$
Hom(Tor_n(Y,X),Z)\cong Hom(Y,Ext^n(X,Z))?
$$
In my head I am thinking this could be useful as follows:
For any two abelian groups $A$ and $B$, $Tor_n^\mathbb{Z}(A,B)=0$ for $n\geq 2$. This is easy to see by taking a projective resolution $0\to ker(f)\to F\to A\to 0$, where $f:F\to A$ is a realization of $A$ as a quotient of a free group $F$. 
From this, can we immediately conclude $Ext^n(A,B)=0$ for $n\geq 2$, without doing anymore work? (I am sure a direct proof would be equally straightforward, I am just interested in the hypothetical). 
Thanks in advance
 A: No, here is why.
Recall that a right adjoint is left exact. So if $\operatorname{Ext}^n(X,.)$ were the right adjoint of some functor, it should preserve left exact sequence. But it doesn't ! Indeed, if $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence, then the long exact sequence
$$...\rightarrow\operatorname{Ext}^{n-1}(X,C)\rightarrow\operatorname{Ext}^n(X,A)\rightarrow\operatorname{Ext}^n(X,B)\rightarrow\operatorname{Ext}^n(X,C)\rightarrow ...$$
shows that in general the map $\operatorname{Ext}^n(X,A)\rightarrow\operatorname{Ext}^n(X,B)$ is not into.
But the tensor-hom adjunction leads indeed to a derived version. For this, we need a more advanced tool : the derived category. (For simplicity, I assume that we are in $R$-Mod, moreover there are technical boundedness assumptions that I will not expand here). There are complexes $Y\otimes^L X$ and $R\operatorname{Hom}(X,Z)$, such that $$\operatorname{Hom}_D(Y\otimes^L X,Z)=\operatorname{Hom}_D(Y,R\operatorname{Hom}(X,Z))$$
with $H_n(Y\otimes^L X)=\operatorname{Tor}_n(Y,X)$ and $H^n(R\operatorname{Hom}(X,Z))=\operatorname{Ext}^n(X,Z)$.
I am not sure we can deduce formally the vanishing of $\operatorname{Ext}^2_\mathbb{Z}$ from the above derived adjunction (I will think about it). But if you know that you can compute $\operatorname{Ext}$ using a projective resolution of the first variable, the proof for $\operatorname{Tor}$ works for $\operatorname{Ext}$ and many other functors (namely every right exact covariant or left exact contravariant).
