(i)
The hybridisation depends on how hard the ligand pushes on the metal's d-orbitals. Chloride is a weak-field ligand, so it leaves nickel's 3d electrons alone; nickel then mixes one 4s and three 4p orbitals, making $\mathrm{[NiCl_4]^{2-}}$ an $\mathrm{sp^3}$ complex. Cyanide is a strong-field ligand, so it forces nickel's 3d electrons to pair up and frees one 3d orbital; this 3d together with the 4s and two 4p orbitals gives $\mathrm{[Ni(CN)_4]^{2-}}$ a $\mathrm{dsp^2}$ set.
(ii)
The labels inner and outer simply tell us whether an inner 3d orbital was used. $\mathrm{[Ni(CN)_4]^{2-}}$ uses an inner 3d orbital in its $\mathrm{dsp^2}$ set, so it is the inner-orbital complex. $\mathrm{[NiCl_4]^{2-}}$ uses only the outer 4s and 4p orbitals, so it is the outer-orbital complex.
(iii)
Nickel here is $\mathrm{Ni^{2+}}$ with a $\mathrm{3d^8}$ arrangement. In the chloride complex the weak field lets those eight electrons keep two of them unpaired, so it is attracted by a magnet (paramagnetic). In the cyanide complex the strong field squeezes the electrons together until none are left unpaired, so it is pushed away by a magnet (diamagnetic).
So chloride gives sp³ / outer / paramagnetic, and cyanide gives dsp² / inner / diamagnetic.