2. In a regular tetrahedron $ABCD$, the line segment connecting the midpoint of edge $AD$ and the center of face $BCD$ is $MN$, and the line segment connecting the midpoint of edge $CD$ and the center of face $ABC$ is $PQ$. Find the angle between $MN$ and $PQ$.
2. Let the edge of the tetrahedron be \( a \). On the line segment \( AM \), take the segment \( ME = \frac{\alpha}{3} \), and construct the altitude \( DF \) of face \( BCD \) and the altitude \( AF \) of face \( ABC \), as shown in Figure 1511. Then, \( FN = \frac{1}{3} FD \), \( FQ = \frac{1}{3} FA \), \( NQ \parallel AD \), and \( NQ = \frac{1}{3} AD = \frac{\theta}{3} \). Therefore, \( MNQE \) is a parallelogram, so \( EQ \parallel MN \).

Also, \( MN = PQ = \frac{\theta}{2} \). In \( \triangle DPE \), \( DP = \frac{a}{2} \), \( DE = \frac{5}{6} a \), and \( \angle PDE = 60^\circ \), so \( EP^2 = \frac{19}{36} a^2 \). In \( \triangle PQE \), by the cosine rule, we get \( \cos \angle PQE = -\frac{1}{18} \), hence the angle between \( PQ \) and \( MN \) is \( \arccos \frac{1}{18} \).