Solution:

(a) $4 x+8 y+z-8=0$ and $y+z-4=0$
It is given that
The eq. of the given planes are
$4 x+8 y+z-8=0 \text { and } y+z-4=0$
It is known to us that, two planes are $\perp$ if the direction ratio of the normal to the plane is
$\begin{array}{l}
a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\
0+8+1 \\
9 \neq 0
\end{array}$
As a result, both the planes are not $\perp$ to each other.
Let’s now check, two planes are || to each other if the direction ratio of the normal to the plane is
$\begin{array}{l}
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\
\frac{4}{0} \neq \frac{8}{1} \neq \frac{1}{1}
\end{array}$
As a result, the the planes are not $\|$ to each other.
Now find the angle between them which is given as
$\begin{array}{l}
\cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right| \\
\cos \theta=\frac{4 \times 0+8 \times 1+1 \times 1}{\sqrt{16+64+1} \sqrt{0+1+1}} \\
=\frac{9}{9 \sqrt{2}} \\
\theta=\cos ^{-1} \frac{9}{9 \sqrt{2}} \\
=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
=45^{\circ}
\end{array}$
Therefore, the angle is $45^{\circ}$.