Is it true that under certain conditions,$Mg$ can reduce $Al_2O_3$ and $Al$ can reduce $MgO$? What are those conditions?

(N/A) Yes,it is true. This can be explained using the Ellingham diagram,which plots the standard Gibbs free energy of formation $(\Delta_f G^\Theta)$ versus temperature $(T)$.
$1$. The reaction $3Mg + Al_2O_3 \to 3MgO + 2Al$ occurs when the $\Delta_f G^\Theta$ curve for $MgO$ lies below the $\Delta_f G^\Theta$ curve for $Al_2O_3$. This happens at temperatures below $1623 \ K$.
$2$. The reaction $2Al + 3MgO \to Al_2O_3 + 3Mg$ occurs when the $\Delta_f G^\Theta$ curve for $Al_2O_3$ lies below the $\Delta_f G^\Theta$ curve for $MgO$. This happens at temperatures above $1623 \ K$.
$3$. The intersection point of the two curves is $1623 \ K$,where $\Delta_f G^\Theta$ for both reactions is equal.