jmc

The tetrahedron is shown below. In order to find $\cos \angle AMB$, we build a right triangle with $\angle AMB$ among its angles. The foot of the altitude from $A$ to face $BCD$ is the centroid, $G$, of triangle $BCD$.



Since $\overline{BM}$ is a median of $△BCD$, point $G$ is on $\overline{BM}$ such that $GM=\frac{1}{3}BM$. Furthermore, we have $AM=BM$, so $$\cos \angle AMB=\cos \angle AMG=\frac{GM}{AM}=\frac{(BM/3)}{BM}=\boxed{\frac{1}{3}}.$$