A correct statement is

Option (1) $\left[\mathrm{MnBr}_{4}\right]^{2-}$ $\mathrm{Br}^{-}$weak ligand does not allow pairing of $e^{-} s$, therefore, hybridization of the Mn in complex is sp3 and geometry is tetrahedral. Option $(2)\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ $\mathrm{NH}_{3}$ is a strong field ligand, therefore, pairing of electrons will take place. But due to insufficiency of vacant orbital, pairing will not take place Electronic configuration of $\mathrm{Ni}(\mathrm{Z}=8)[\mathrm{Ar}] 3 d^{8} 4 s^{2}$ Electronic configuration of $N i^{2+}$ is $[\mathrm{Ar}] 3 d^{8}$ $N i^{2+}$ in the presence of strong field ligand \begin{array}{|l|l|l|l|l|} \hline 1 L & 1 L & 1 L & 1 & 1 \\ \hline \end{array}
It is outer orbital complex and not an inner orbital complex Option (3) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ $\mathrm{NH}_{3}$ is a strong field ligand, therefore, pairing will take place Electronic configuration of $C_{0}(Z=27)$ is $[A r] 3 d^{74} S^{2}$ Electronic configuration of $\mathrm{Co}^{2+}$ is $[\mathrm{Ar}] 3 d^{7}$ $\mathrm{Co}^{2+}$ in the presence of strong field ligand.
There is one unpaired $e^{-}$therefore, it is paramagnetic Option (4) $\left[\mathrm{CoBr}_{2}(e n)_{2}\right]^{-}$does not have ambidentate ligands, therefore, is does not exhibits linkage isomerism.