In $\triangle ABC$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle AGE$ is equilateral. What is $\cos(B)$?

Let $AE=EG=AG=2$. Since $G$ is the centroid, $DG=\frac12AG=1$ and $BG=2GE=4$; since $E$ is the midpoint of $AC$, $CE=2$ and $AC=4$. In $\triangle AGB$, $\angle AGB=180^\circ-60^\circ=120^\circ$, so $AB^2=4+16+8=28$, $AB=2\sqrt7$. In $\triangle BGD$, $\angle BGD=\angle AGE=60^\circ$, so $BD^2=1^2+4^2-2(1)(4)\cos60^\circ=13$, and $BC=2BD=2\sqrt{13}$.
By the law of cosines in $\triangle ABC$:
$$\cos B=\frac{AB^2+BC^2-AC^2}{2\cdot AB\cdot BC}=\frac{28+52-16}{2(2\sqrt7)(2\sqrt{13})}=\frac{64}{8\sqrt{91}}=\boxed{\frac{8\sqrt{91}}{91}}.$$