There is no orbital hybridization in either of these diatomic compounds.
Note that in linear diatomic molecules, the pz orbital always points along the internuclear axis, so it has to contribute to one of the σ bonds.
I've drawn the overlaps below in the MO diagrams.
BROMINE BONDING (HOMONUCLEAR DIATOMIC)
For Br2, it is the simpler of the two examples. It is a homonuclear diatomic, so all of its orbitals have a compatible partner: 4px with 4px, 4py with 4py, etc.
In this case, since there are only two atoms, they use their highest energy compatible orbitals to σ bond on the internuclear axis. The energy of the 4s atomic orbital is −24.37 eV, and the 4p atomic orbitals are −12.49 eV in energy (Inorganic Chemistry, Miessler et al., Table 5.2).
Each bromine would donate one 4pz electron to form a σ-bonding orbital.
As a result, there is no orbital hybridization here.
Here is the MO diagram below (I had to draw it myself since I couldn't find one online; the π4px and π4py orbitals---the 1b3u and 1b2u---are very close in energy to the σ4pz, the 1b1u):
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Granted, that is not Br2's highest-occupied molecular orbital (that would be the π*4px and π*4py---the 1b3u and 1b2u), but since both the bonding and antibonding π molecular orbitals are occupied, it is the σ4pz (1b1u) that participates in the σ bond.
You should notice that the 1b1u orbital is the σ4pz bonding orbital, but the 2b1u---the σ*4pz antibonding orbital---has no electrons, so it doesn't contribute to the σ bond. If it did, Br2 would not exist.
Therefore, the σ4pz indeed is the molecular orbital that represents the single bond on Br2.
NO+ BONDING (HETERONUCLEAR DIATOMIC)
NO+, on the other hand, is a heteronuclear diatomic. Since it is also diatomic, it also does not need to hybridize.
All of nitrogen's orbitals are compatible with oxygen's orbitals in energy (and in symmetry, but that is less crucial to our understanding for General Chemistry level education).
The MO diagram for neutral NO is as follows (Inorganic Chemistry, Miessler et al., Ch. 5, Answer Key):
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(I superimposed some orbital depictions on the original diagram, and added symmetries and energies.)
If we consider NO+, we remove the electron from the highest-occupied molecular orbital, so we take out the one from the π*2px antibonding orbital (2b1) to form NO+.
At this point, its bonds have increased in strength. The bond order changed from:
8−32=2.5
to:
8−22=3
So we know it has a triple bond. That means it needs three orbitals contributed from each atom.
There are two electrons in the σ2pz molecular orbital (3a1), and there are two electrons each in the π2px (1b1) and π2py (1b2) molecular orbitals.
NO+ therefore uses two 2px atomic orbitals, two 2py atomic orbitals, and two 2pz atomic orbitals to bond.
As a result, there is no orbital hybridization here.
Each σ-bonding pair contributes to a σ bond, and each π-bonding pair contributes to a π bond. That accounts for the triple bond: one σ and two π bonds.
