This problem includes conceptual mixing of crystal field theory and magnetic moment $(\mu)$ determination.
(a) $[CoF_6]^{3-}$.
$F^{-}$is a weak field ligand.
Configuration of $Co^{3+}=3 d^{6}(.$ or $.t _{2 g}^{4} e_g^{2})$
Number of unpaired electrons $(n)=4$
Magnetic moment $(\mu)=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.9 BM$
$[Co(H_2 O)_6]^{2+}$,
$H_2 O$ is a weak field ligand.
Configuration of $Co^{2+}=3 d^{7}(.$ or $.t _{2 g}^{5} e_g^{2})$
Number of unpaired electrons $(n)=3$
$\mu=\sqrt{3(3+2)}=\sqrt{15}=3.87 BM$
$[Co(CN)_6]^{3-}$ i.e., $Co^{3+}$
$\because CN$ is strong field ligand.
$Co^{3+}=3 d^{6}(.$ or $.t _{2 g}^{6} e_g^{0})$
There is no unpaired electron, so it is diamagnetic.
$
\mu=0
$
(b) $[FeF_6]^{3-}$,
Number of unpaired electrons, $n=5$
$
\begin{aligned}
\mu & =\sqrt{5(5+2)} \\\
& =\sqrt{35}=5.92 BM
\end{aligned}
$
$[Fe(H_2 O)_6]^{2+}$
$
Fe^{2+}=3 d^{6}(\text{ or } t _{2 g}^{4} e_g^{2})
$
Number of unpaired electrons, $n=4$
$
\begin{aligned}
\mu & =\sqrt{4(4+2)} \\\
Z & =\sqrt{24} \\\
& =4.98 BM
\end{aligned}
$
$[Fe(CN)_6]^{4-}$
Since, $CN^{-}$is a strong field ligand, all the electrons get paired.
$Fe^{2+}=3 d^{6}(.$ or $.t _{2 g}^{6} e_g^{0})$
Because there is no unpaired electron, so it is diamagnetic in nature.