$A$ solution of $KMnO_4$ on reduction yields either a colourless solution,a brown precipitate,or a green solution depending on the $pH$ of the solution. What different stages of the reduction do these represent and how are they carried out?

The reduction of $KMnO_4$ depends on the $pH$ of the medium:
$1$. In acidic medium $(pH < 7)$: $MnO_4^-$ is reduced to $Mn^{2+}$ (colourless solution). The reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
$2$. In neutral or faintly alkaline medium $(pH \approx 7)$: $MnO_4^-$ is reduced to $MnO_2$ (brown precipitate). The reaction is: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$.
$3$. In strongly alkaline medium $(pH > 7)$: $MnO_4^-$ is reduced to $MnO_4^{2-}$ (green solution). The reaction is: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$.