How to make commutative algebraic groups strongly dualizable? Let's use the notation of [A=>B] for Hom(A, B). Take a 1-dimensional algebraic torus Gm and higher-dimensional torus T and let's live in the category of commutative algebraic groups over k.
Out of  four expressions like [Gm=> [Gm=>T]] etc. half give back T (*), others the dual torus TV, in the sense that X*(T) := [Gm => T] = [TV => Gm] =: X*(TV).
The equality  (*) can be proven by using the following formula with B = Gm
   (**)              A \otimes [B=>B] ==== [[A=>B] => B].


Question:  Is there another example of commutative algebraic group or a similar generalized object B, for which the identity (**) is true or true in some generalized sense?

One thing I specifically have in mind is that if we could write [X => Y] = X* \otimes Y whenever X and Y are groups, as if they were vector spaces, the formula would hold for all A and B. So, what's a category that is related to algebraic groups but which posesses this property?
 A: The formulas you are writing seem to arise from Cartier duality, which gives a collection of antiequivalences of categories of group schemes and sheaves of various types.  For the case of tori, one gets a correspondence with (fpqc sheaves of) finitely generated free abelian groups.  You can find some of this in SGA3 volume 1 attached to phrases like "diagonalizable groups" and "groups of multiplicative type".  Gm has a distinguished role as the dualizing object.  It is Cartier dual to the constant sheaf Z.  I don't think there is a replacement group B that satisfies all of the identities you want, because the Cartier dual of B won't be suitably universal in the category of (sheaves of) abelian groups.
A: Since Ilya asked I'll write out in a bit more detail the case of finitely generated abelian groups. There is not (as far as I know) any reasonable notion of strong duality on the abelian category of finitely generated abelian groups in the sense you ask for. Indeed, Z is the tensor unit and any torsion group gets killed by Hom(-,Z). But if one passes to D^b(Z-mod) one does get a rigid tensor category. Setting D(-) = RHom(-,Z) one gets that D(A)\otimes^L B \iso RHom(A,B), so in particular D^2 \iso Id and with the suitably derived tensor and internal hom D^b(Z-mod) is a rigid tensor category. It even turns out that this structure is enough to canonically recover Spec Z as a locally ringed space. As Scott pointed out in his comment one does need finiteness conditions for this to work, the full unbounded derived category of all abelian groups is only a closed tensor category.
In fact this story is true for D^{perf}(X) for any quasi-compact quasi-separated scheme X. The finiteness conditions being necessary is also quite general - in good cases for compactly generated triangulated categories with a tensor product (which respects the triangulation and coproducts) one always gets an internal hom and often the compacts form a rigid tensor category (one can give conditions under which this is guaranteed).
For a category of actual algebraic groups with this property I'd recommend having a look at the category of finite flat commutative group schemes over some base S. Cartier duality gives an honest duality of this category with itself and maybe at least for suitably good S one might get further good properties?
