jmc

The tetrahedron is shown below. In order to find $\cos \angle ABM$, we build a right triangle with $\angle ABM$ 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 $BG=\frac{2}{3}BM$. From 30-60-90 triangle $BMC$, we have $BM=\frac{\sqrt{3}}{2}\cdot BC$, so $$BG=\frac{2}{3}BM=\frac{2}{3}\cdot \frac{\sqrt{3}}{2}\cdot BC=\frac{\sqrt{3}}{3}\cdot BC.$$Finally, since $AB=BC$, we have $$\cos \angle ABM=\cos \angle ABG=\frac{BG}{AB}=\frac{(\sqrt{3}/3)BC}{BC}=\boxed{\frac{\sqrt{3}}{3}}.$$