jmc

Since $AD$ is a median, $D$ is the midpoint of $BC$, so $BD=CD=4$. Let $P$ be the projection of $A$ onto $BC$. (Without loss of generality, we may assume that $P$ lies on $BD$.) Let $x=BP$, so $PD=4-x$. Let $h=AP$.



Then by Pythagoras on right triangles $APB$, $APC$, and $APD$, $$\begin{array}{rl}A{B}^2& =x^2+h^2,\\ A{C}^2& =(8-x{)}^2+h^2,\\ 25& =(4-x{)}^2+h^2.\end{array}$$Adding the first two equations, we get $$A{B}^2+A{C}^2=x^2+h^2+(8-x{)}^2+h^2=2x^2-16x+64+2h^2.$$But from the third equation, $25=x^2-8x+16+h^2$, so $$\begin{array}{rl}A{B}^2+A{C}^2& =2x^2-16x+64+2h^2\\ & =2(x^2-8x+16+h^2)+32\\ & =2\cdot 25+32\\ & =82.\end{array}$$Hence, from the given data, $A{B}^2+A{C}^2$ can only take on the value 82. Therefore, $M=m=82$, so $M-m=\boxed{0}$.