$\left[\mathrm{CoF}_6\right]^{3-} \Rightarrow \mathrm{Co}^{3+}[\mathrm{Ar}] 3 \mathrm{~d}^6 45^{\circ}$
$\mathrm{Sp}^3 \mathrm{~d}^2$ Hybridisation, Octahedral shape
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \Rightarrow \mathrm{Co}^{3+}[\mathrm{Ar}] 3 \mathrm{~d}^6 45^{\circ}$
$\mathrm{d}^2 \mathrm{sp}^3$ Hybridisation, pairing due to strong field ligand
In $\left[\mathrm{NiCl}_4\right]^{2-} \mathrm{Ni}$ is in +2 oxidation state $\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^8$
Hybridization involved is $\mathrm{sp}^3$ as participating orbitals are $4 \mathrm{~S} \& 4 \mathrm{P}$.
In $\left[\mathrm{NiCN}_4\right]^{2-} \mathrm{Ni}$ is in +2 oxidation state.
$\mathrm{Ni}^{+2}-[\mathrm{Ar}] 3 \mathrm{~d}^8$
$\mathrm{CN}^{-}$is strong field ligand. It causes pairing of $3 \mathrm{~d}$ electrons.
Hybridization involved is $\mathrm{dsp}^2$ as participating orbitals are $3 \mathrm{~d}$, $4 \mathrm{~S}$ and $4 \mathrm{P}$. Compound is diamagnetic due to pairing of electrons and shape is square planar.