7. For a regular tetrahedron \(ABCD\), the center of the circumscribed sphere is \(O\), and \(E\) is the midpoint of edge \(BC\). The dihedral angle \(A-BO-E\) is \(\qquad\).
II. $7 . \frac{2 \pi}{3}$.
Since point $E$ is on side $B C$, the dihedral angle $A-B O-E$ is equal to the dihedral angle $A-B O-C$. Also, $O$ is the center of the circumscribed sphere of the regular tetrahedron $A B C D$, so $B O \perp$ plane $A C D$. Let the projection of point $B$ on plane $A C D$ be $O^{\prime}$. Then $O^{\prime}$ is the center of the equilateral $\triangle A C D$. Therefore, $B O \perp A O^{\prime}, B O \perp C O^{\prime}$.
Thus, $\angle A O^{\prime} C$ is the plane angle of the dihedral angle $A-B O-C$, and $\angle A O^{\prime} C=\frac{2 \pi}{3}$.