jmc

The tetrahedron is shown below. In order to find $\tan \angle AMB$, 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 $GM=\frac{AM}{3}$. The Pythagorean Theorem gives us $$AG=\sqrt{A{M}^2-G{M}^2}=\sqrt{A{M}^2-\frac{A{M}^2}{9}}=AM\cdot \sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}\cdot AM.$$ Finally, we have $$\tan \angle AMB=\tan \angle AMG=\frac{AG}{GM}=\frac{(2\sqrt{2}/3)AM}{AM/3}=\boxed{2\sqrt{2}}.$$