 Solution

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence