Is Hilbert Stress-energy tensor always same as Belinfante stress-energy tensor?
Well, firstly, we have to assume Lorentz covariance and general covariance of the theory. For non-relativistic theories all bets are off. Secondly, in case of fermions, one needs to generalize the Hilbert SEM tensor from variation wrt. a metric to a variation wrt. a vielbein, see e.g. my Phys.SE answer here. Then the generalized Hilbert SEM tensor is the canonical SEM plus the Belinfante-Rosenfeld improvement term. A proof is sketched in my Phys.SE answer here and links therein.
No, a symmetric SEM tensor is not unique. It is in principle possible to add improvement terms that respects the symmetry and the conservations laws.
Yes, provided we work with Torsion-free connections. This is explained in the original papers of Belinfante and Rosenfeld that are cited on the Wikipedia page.
You can see the torsion-free necessity by using a vierbein formulation. Just varying the vierbein alone usually gives the asymmetric Noether tensor. If you define the spin current by the variation of the spin connection, and link the spin connection to the metric by the torsion-free condition, then the spin-connection variation generates precisely the extra Belinfante-Rosenfeld terms.