Are properties of Tor, Ext analogous to those of tensor product and Hom in the following two ways?

$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\Ext}{\operatorname{Ext}}$$\newcommand{\Tor}{\operatorname{Tor}}$ Questions:

1. While the contravariant $\Ext$ functor is the first derived functor of the contravariant $\Hom$ functor, is the covariant $\Ext$ functor the first derived functor of the covariant $\Hom$ functor?

2. Does the fact that $- \otimes X \simeq X \otimes -$, both covariant functors, imply in turn that: $$\Tor(-,X) \simeq \Tor(X,-),\text{ both covariant?}$$

Yes/no answers will suffice, although if an answer is no, I would also appreciate a suggestion for a reference that discusses these issues further, or even your own explanation if you feel up to it.

Background Information: Please correct me if I am wrong -- my current understanding is:

1. The tensor product functor $- \otimes B$ and the co-variant Hom functor $\Hom(B, -)$ are an adjoint pair (tensor the left adjoint, co-variant Hom functor the right adjoint).

2. Any functor which is a left adjoint is right exact, and any functor which is a right adjoint is left exact. From 1., it follows as a special case that the tensor product functor is right exact and the co-variant Hom functor is left exact.

3. Because the tensor product functor is right exact but not left exact, we can use this "lack of left exactness" to define/form its first derived functor, the Tor functor.

4. Because the contra-variant Hom functor is left exact but not right exact, we can use this "lack of right exactness" to define/form its first derived functor, "the" Ext functor.

To clarify, I know 1. and 2. from my commutative algebra course, and 3. and 4. from my homology course. Since the two courses don't intersect or interact, I have to clarify any connections myself.

Consulting this page, I was reminded that one could define at least two different Ext functors: $\operatorname{Ext}(-,X)$ and $\operatorname{Ext}(X,-)$. Only the first one is derived from the contra-variant Hom functor (I think), and that page says that it is contravariant, while the other is covariant.

Corresponding to this, we could define at least two different Tor functors, $\operatorname{Tor}(X,-)$ or $\operatorname{Tor}(-,X)$. However, if the two tensor product functors are isomorphic and both covariant, one would expect (naively) that these two are also isomorphic and both covariant.

1. $$\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Ext}{\operatorname{Ext}}\newcommand{\Tor}{\operatorname{Tor}}$$Yes, it is. For any $$i\geq 0$$ you can compute $$\mathrm{Ext}^i(A,B)$$ both as the $$i-th$$ derived functor of $$\Hom(A,-)$$ and as the $$i-th$$ derived functor of $$\Hom(-,B)$$. Of course the first is a right derived functor (of a left exact functor) while the second is a left derived functor (of a right exact functor). This result is called balancing of the Ext functor. Notice that this implies in particular that $$\Ext^i(A,B)=0$$ whenever $$A$$ is projective or $$B$$ is injective.
2. In general, if $$F$$ and $$G$$ are two functors, both covariant (or both contravariant) and they are isomorphic, then their derived functors are all equal, in symbols $$R^iF \simeq R^iG$$ for every $$i\geq 0$$. This follows directly from the construction of the derived functors. In particular, since $$-\otimes X$$ and $$X\otimes -$$ are isomorphic covariant functors, $$\Tor^i(-,X)\simeq \Tor^i(X,-)$$.