2. 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. $\pi-\arctan \frac{\sqrt{2}}{2}$
2. D As shown in the figure, construct $E E_{1} \perp$ plane $B C D$ at $E_{1}$, then $E_{1}$ is on $B G$;

Since $A B C D$ is a regular tetrahedron, $A C \perp B D$.
Also, because $E F / / A C, F G / / B D$, it follows that $E F \perp F G$, hence $E_{1} F \perp F G$.
Therefore, $\angle E F E_{1}$ is the plane angle of the dihedral angle $E-F G-B$, denoted as $\alpha$.
Let the edge length of the regular tetrahedron $A B C D$ be 1, then its height $A H=\sqrt{A B^{2}-B H^{2}}=\sqrt{1^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\frac{\sqrt{6}}{3}$.
Thus, $E E_{1}=\frac{1}{2} A H=\frac{\sqrt{6}}{6}$, and $E F=\frac{1}{2} A C=\frac{1}{2}, \alpha=\arcsin \frac{\frac{\sqrt{6}}{6}}{\frac{1}{2}}=\arcsin \frac{\sqrt{6}}{3}$,
Therefore, the size of the dihedral angle $C-F G-E$ is $\pi-\arcsin \frac{\sqrt{6}}{3}=\pi-\arctan \sqrt{2}$.