a) For the given complex \[\left[ {Co{{(N{H_3})}_5}Br} \right]S{O_4}\], write the IUPAC name and its ionization isomer.
b) Which sets of d-orbitals of metal ion/atom experience more repulsion in the octahedral field created by the ligand?

Hint: For writing name of the complex we must know the IUPAC nomenclature and naming of ligands. We can get ionization isomers by exchanging the number of ions between the coordination sphere and counter ion.

Complete step by step answer:
Let us start by writing the IUPAC nomenclature, the \[{\text{N}}{{\text{H}}_3}\] group is names as ammine, the \[{\text{B}}{{\text{r}}^ - }\] is called bromo and \[{\text{SO}}_4^{2 - }\] as sulphate. The oxidation state of cobalt here is . So IUPAC name will be: pentamminebromocobalt(III) sulphate.
Ionization isomers are those which have the same molecular formula but give different ions when ionized, but the oxidation state of metal must not change. Since ammonia is a neutral ligand having zero charge and sulphate have negative charge. So changing them will change the oxidation state of metal ions. So we will exchange the sulphate ion with bromide ion. Hence the ionization isomer will be, \[\left[ {Co{{(N{H_3})}_5}S{O_4}} \right]Br\].
Broadly there are two sets of d orbitals, axial and nonaxial. Axial orbital’s are along the axes which are \[{d_{{x^2} - {y^2}}}{\text{ and }}{d_{{z^2}}}\]. In octahedral complexes ligand approaches through axial direction and ligand carry charge or electron density and hence cause repulsion with electron density of metal ion because same charges repel. Due to this reason the axial orbitals experience more repulsion.

Note:
Tips for writing the nomenclature, always ligands are named first in complexes. The cationic part is written first followed by anionic part. Leave a space between the name of cation and anion. If oxidation states are to be written it is written in parenthesis along with metal ion roman numbers. Prefixes are written in case of a number of ligands repeating such as bi, tri, penta etc.