1) $2x + y + 3z = -5$ and $x - y + z = 1$
solution:
$$[2,1,3] \cdot [1,-1,1] = 4$$
$$\|a\| \|b\| = \sqrt{14}\sqrt{3}$$
$$\theta = cos^{-1}\left(\frac{4}{\sqrt{14}\sqrt{3}} \right) = 51.89$$
2) $6x + 14y -2z = 4$ and $-9x -21y +3z = -1$
Solution:
$$[6,14,-2] \cdot [-9,-21,3] = -354$$
$$\|a\| \|b\| = \sqrt{236}\sqrt{531}$$
$$\cos^{-1}\left(\frac{-354}{\sqrt{236}\sqrt{531}}\right) = 180$$
Is this correct?
Yes that's correct, for the second note that
$$[6,14,-2] =-\frac23 [-9,-21,3]$$
and the two planes are defined parallel. Using angles, note that usually we refer to $\theta \le 90°$ and thus $\theta = 0°$ is maybe preferable with respect to $180°$, which refers to the specific orientation of the selected normal vectors.