5. Let $E, F, G$ be the midpoints of the edges $A B, B C, C D$ of a regular tetrahedron $A B C D$, respectively. Then the dihedral angle $C-F G-E$ is
(A) $\arcsin \frac{\sqrt{6}}{3}$
( B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$.
(C) $\frac{\pi}{2}+\arctan \sqrt{2}$
(D) $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$
(D)
5.【Analysis and Solution】As shown in the figure, draw $E E_{1} \perp$ plane $B C D$, then $E E_{1} \perp B G$.
$\because A B C D$ is a regular tetrahedron,
$\therefore A C \perp B D$. Also, $\because E F / / A C, F G / / B D, \therefore E F \perp F G$.
Thus, $E_{1} F \perp F G$ so $\angle E F E_{1}$ is the plane angle of the dihedral angle $E-F G-B$, denoted as $a$.
Assume the edge length of the regular tetrahedron $A B C D$ is 1, then its height $A H=\sqrt{A B^{2}-B H^{2}}=\sqrt{1^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\frac{\sqrt{6}}{3}$.
$$
\begin{array}{l}
\therefore E E_{1}=\frac{1}{2} A H=\frac{\sqrt{6}}{6} . \\
\text { Also, } E F=\frac{1}{2} A C=\frac{1}{2}, \\
\therefore a=\arcsin \frac{\frac{\sqrt{6}}{6}}{\frac{1}{2}} \\
=\arcsin \frac{\sqrt{6}}{3} .
\end{array}
$$

Therefore, the size of the dihedral angle $O-F G-E$ is
$$
\pi-\arcsin \frac{\sqrt{6}}{3}=\pi-\operatorname{arccot} \frac{\sqrt{2}}{2} \text {. }
$$