(a) Determine the symmetries (irreducible components) of the MOs responsible for the d-d metal-metal quadruple bonding in this molecule
I'll try to walk through how it works for $d_{xy}$. You will need to consider $d_{yz}$ and $d_{xz}$ as a pair later when you attempt those.

Obviously $\chi(E)=2$. Now for the $C_4$ rotation, just think about rotating one of the orbitals in the $xy$ plane (the same thing will happen to the other orbital, so we will then double the character). When we do this, the orbital remains in place, but it's positive and negative lobes invert. So it will contribute -1 and for both orbitals $\chi(C_4)=-2$. For $C_2$, the rotation again leaves the orbital in place and leaves the lobes in each location with the same sign so $\chi(C_2)=2$. 

$\chi(C_2')=\chi(C_2'')=\chi(i)=\chi(S_4)=\chi(\sigma_h)=0$ since all of these operations involve a rotation/reflection around a parallel line/plane so they will move the orbitals. 

$\sigma_v$ runs along the $\ce{Re-Cl}$ bonds. Along and on either side of this plane, the sign of the lobes is the same and so $\chi(\sigma_v)=2$. $\sigma_d$ runs between the $\ce{Re-Cl}$ bonds and so this plane inverts the sign and $\chi(\sigma_d)=-2$.

Plugging these into a calculator I get a $b_{1g}$ and a $b_{2u}$ irrep. I will leave it to you to figure out which of these is the bonding combination and which is antibonding.