What kind of algebraic object is $\mathcal{D}_X$? (algebra of diifferential operators). What's special about modules over it? Let $R$ be a regular ring over a field of char 0. Let $X=Spec R$ and $D=\mathcal{D}_X$
the algebra of differential operators over it. 
The overall vague question is what kind of algebraic object is $D$ and what kind of category is the category of its modules? Here are some points whose answers could together be considered an answer to this.

  
*
  
*When (meaning for what kind of $D$-modules $M,N$ left or right) is there a $D$-module structure on $Hom_R(M,N)$? On $M\otimes_RN$? Does this make
  $D$-mod into an abelian monoidal category with fiber functor to
  $R$-mod? If so is it closed monoidal?
  
*When does a $D$-module $M$ admit a dual $M^*$? (in the sense of monoidal categories).
  
*Is the abelian category of $D$-modules isomorphic to $D^{op}$-mod? (before deriving). If not how are they related? Maybe the correct
  thing to consider is the opposite co-opposite $D^{op}_{cop}$? Specifically Why does the dualizing sheaf $\omega_X$ pop up in this context?
  
*Is the $R$-linear dual $Hom_R(D,R)$ a bialgebra? Is it related to the opposite of $D$?
  
*How do I derive correctly the category of $D$-modules? Suppose $M$ and $N$ are $D$-modules and suppose $Hom_R(M,N)$ is the "correct" internal Hom in the abelian monoidal category of $D$-modules. If we hope to have a fiber functor between the derived categories $Hom_R(Q,N)$ (with $Q \to M$ a resolution as a $D$-module) should go to (something quasi isomorphic) to $Hom_R(P,N)$ (with $P \to M$ a resolution as an $R$-module). It doesn't seem to follow easily from the rest of the structure.

 A: *

*Proposition 1.2.9 of http://math.columbia.edu/~scautis/dmodules/hottaetal.pdf explains that if $M$ and $N$ are both left $D$-modules and $M'$ and $N'$ are both right $D$-modules then
(a) $M\otimes_{R} N$ is naturally a left $D$-module;
(b) $M'\otimes_{R} N$ is naturally a right $D$-module;
(c) $\mathrm{Hom}_{R}(M,N)$ is naturally a left $D$-module;
(d) $\mathrm{Hom}_{R}(M',N')$ is naturally a left $D$-module;
(e) $\mathrm{Hom}_{R}(M,N')$ is naturally a right $D$-module.
This Proposition also gives explicit formulae that explain why each of these is the case. Remark 1.2.10 explains why $M'\otimes_{R}N'$ is not naturally a left or right $D$-module in general. 
Section 2 of https://arxiv.org/abs/dg-ga/9702008 fits these results into a wider framework. The thing about $D$ that makes these things true is that it is the universal enveloping algebra of a $(k,R)$ Lie-Rinehart algebra. 
I think that it is clear from what I've written that $D$-mod (=cat of left $D$-modules) is a (symmetric) monoidal category with $\otimes=\otimes_{R}$ and  the forgetful functor $D$-mod to $R$-mod preserves the monoidal product. I'm not sure if a fiber functor requires more than this. 
As noted by t3suji in the comments below the internal Hom $\mathrm{Hom}_R(-,-)$ on $R$-mod which induces an internal Hom on $D$-mod via part (c) above does make $D$-mod into a closed monoidal category. Once again this observation works for the enveloping algebra of any Lie-Rinehart algebra. 

*I think it is probably the case that $M$ has a dual precisely if it is a projective $R$-module of finite rank (that is also a left $D$-module). In this case $M^\ast$ should just be $\mathrm{Hom}_R(M,R)$ which is a left $D$-module by 1(c). 

*Yes. The reason that $\omega_X$ crops up is that it is a right $D$-module that is a rank $1$ projective $R$-module. It follows that tensoring with it defines an auto-equivalence of categories on $R$-mod (the inverse is given by $\mathrm{Hom}_R(\omega_X,-)\cong \mathrm{Hom}_R(\omega_X,R)\otimes_R-$. It follows from 1(b) that this auto-equivalence sends left $D$-modules to right $D$-modules and from 1(d) that its inverse sends right $D$-modules to left $D$-modules. 

*Since $D$ is not a finitely generated $R$-module, $\mathrm{Hom}_R(D,R)$ is pretty badly behaved. In particular it is unlikely that the algebra structure on $D$ will induce a coalgebra structure on its 'dual'. Incidentally I'm not sure what you mean by saying that $D$ is an $R\otimes R$ bialgebra (in particular I don't know what the coalgebra structure is nor the counit). 

*I don't really understand what this part is asking. 
Edited in the light of the comments below by t2suji
