Solution:

(a) $7 x+5 y+6 z+30=0$ and $3 x-y-10 z+4=0$
It is given that
The eq. of the given planes are
$7 x+5 y+6 z+30=0$ and $3 x-y-10 z+4=0$
Two planes are $\perp$ if the direction ratio of the normal to the plane is
$\begin{array}{l}
a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\
21-5-60 \\
-44 \neq 0
\end{array}$
Both the planes are not $\perp$ to each other.
So now, two planes are || to each other if the direction ratio of the normal to the plane is
$\begin{array}{l}
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\
\frac{7}{3} \neq \frac{5}{-1} \neq \frac{6}{-10} \text { [both the planes are not } \| \text { to each other] }
\end{array}$
Therefore, the angle between them is given by,
$\begin{aligned}
\cos \theta =\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right| \\
\cos \theta =\frac{-44}{\sqrt{49+25+36} \sqrt{9+1+100}} \\
=\frac{-44}{\sqrt{110} \sqrt{110}} \\
=\frac{-44}{110} \\
\theta=\cos ^{-1} \frac{2}{5}
\end{aligned}$
As a result, the angle is $\cos ^{-1}(2 / 5)$

(b) $2 x+y+3 z-2=0$ and $x-2 y+5=0$
It is given that
The eq. of the given planes are
$2 x+y+3 z-2=0$ and $x-2 y+5=0$
Two planes are $\perp$ if the direction ratio of the normal to the plane is
$\begin{array}{l}
a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\
2 \times 1+1 \times(-2)+3 \times 0 \\
=0
\end{array}$
As a result, the given planes are $\perp$ to each other.