How to balance this redox reaction taking place in a basic medium?
Writing an answer since I can't comment yet.

You are using the ion-electron method for basic medium, but are balancing using $\ce{H+}$ and later adding $\ce{OH-}$? Sounds suspicious to me, but the outcome must be the same.

Generally, I add $\ce{OH-}$ ions on the oxygen deficient side and then add water to compensate the hydrogen deficiency on the other side.

Note: your first half-reaction balancing the conversion of permanaganate to $\ce{MnO2}$ must be regarded incorrect, as $\ce{MnO2}$, being soluble in acidic conditions (presence of $\ce{H+}$, as the equation indicates!) would yield to further reduction, and end up as $\ce{Mn^2+}$.

Here's how I balanced it:


Oxidation half


$$
\begin{align}
\ce{Al &amp;-&gt; Al(OH)4-}\\
\ce{Al + 4OH- &amp;-&gt; Al(OH)4- + 3e-}
\end{align}
$$


Reduction half:


$$
\begin{align}
\ce{MnO4- &amp;-&gt; MnO2}\\
\ce{MnO4- &amp;-&gt; MnO2 + 2OH-}\\
\ce{MnO4- + 2 H2O &amp;-&gt; MnO2 + 4 OH- + 3 e-}
\end{align}
$$


Adding:


$$
\ce{Al + MnO4- + 2 H2O -&gt; Al(OH)4- + MnO2}
$$ 

Your method is fine as well. The only mistake is the wrong $\ce{H+}$ coefficient in $\ce{Al}$ oxidation half-reaction, which will be 4 instead of 8.

P. S. (only concern): note the differences in half reactions in basic and acidic media. Conclusion: when products differ in acidic/basic conditions, it is best not to balance in the asker's method, as it cannot be considered entirely correct. Nothing more, nothing less.