3. In the tetrahedron $A-B C D$, all edges are of length $1, E, F$ are the midpoints of $A C, A D$ respectively, $O$ is the center of $\triangle B C D$, find the angle formed by the skew lines $B E$ and $F O$.
3. Connect $D O$ intersecting $B C$ at $G$, connect $A G$, intersecting $B E$ at $H$. Since $B G \perp A G, B G \perp D G$, it is known that the projection of $B H$ on plane $A G D$ is $G H$. Therefore, $\cos \theta_{1}=\cos \angle B H G=\cos 60^{\circ}=\frac{1}{2}$. Take the midpoint $M$ of $D G$, connect $F M$, then $F M / / H G$, so $\theta_{2}=\angle O F M$. Connect $A O$, then $A O \perp O D$, hence $O F=\frac{1}{2} A D=\frac{1}{2}, F M=\frac{1}{2} A G=\frac{\sqrt{3}}{4}, O M=O D-\frac{1}{2} D G=\frac{\sqrt{3}}{12}$, so $\cos \theta_{2}=$ $\frac{O F^{2}+F M^{2}-O M^{2}}{2 O F \cdot F M}=\frac{5}{3 \sqrt{3}}$, thus $\cos \theta=\frac{1}{2} \cdot \frac{5}{3 \sqrt{3}}=\frac{5 \sqrt{3}}{18}, \theta=\arccos \frac{5 \sqrt{3}}{18}$.