(7) For a regular tetrahedron \(ABCD\) with the center of its circumscribed sphere being \(O\), and \(E\) being the midpoint of \(BC\), the dihedral angle \(A-BO-E\) is \(\qquad\).
(7) $\frac{2}{3} \pi$ Hint: Since $E$ is on $BC$, the dihedral angle $A-BO-E$ is equal to the dihedral angle $A-BO-C$; since $O$ is the center of the circumscribed sphere of the regular tetrahedron $ABCD$, $BO \perp$ plane $ACD$, let the projection of $B$ on plane $ACD$ be $O'$, then $O'$ is the center of the equilateral triangle $ACD$; $BO \perp AO', BO \perp CO'$, so $\angle AO'C$ is the plane angle of the dihedral angle $A-BO-C$, $\angle AO'C=\frac{2}{3} \pi$.