Solution:

It is known to us that
If
$\frac{\mathrm{x}-\mathrm{x}_{1}}{1_{1}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~m}_{1}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{n}_{1}} \text { and } \frac{\mathrm{x}-\mathrm{x}_{2}}{1_{2}}=\frac{\mathrm{y}-\mathrm{y}_{2}}{\mathrm{~m}_{2}}=\frac{\mathrm{z}-\mathrm{z}_{2}}{\mathrm{n}_{2}} \text { are the eq. of }$ two lines, then the acute angle between the two lines is given by $\cos \theta=\left|l_{1} \mathrm{l}_{2}+\mathrm{m}_{1} \mathrm{~m}_{2}+\mathrm{n}_{1} \mathrm{n}_{2}\right| \ldots \ldots(1)$

(i) $\frac{x-2}{2}=\frac{y-1}{5}-\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}-\frac{z-5}{4}$
Here, $a_{1}=2, b_{1}=5, c_{1}=-3$ and
$a_{2}=-1, b_{2}=8, c_{2}=4$
Now,
$1=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}, \mathrm{~m}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}, \mathrm{n}=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}} \ldots \ldots \text { (2) }$
Here, it is known to us that
$\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}=\sqrt{2^{2}+5^{2}+(-3)^{2}}=\sqrt{4+25+9}=\sqrt{38}$
And
$\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}=\sqrt{(-1)^{2}+8^{2}+4^{2}}=\sqrt{1+64+16}=\sqrt{81}=9$
From eq. (2), we have
$\begin{aligned}
1_{1}=\frac{a_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\frac{2}{\sqrt{38}}, m_{1}=\frac{b_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\frac{5}{\sqrt{38}}, n_{1} =\frac{c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}} \\
=\frac{-3}{\sqrt{38}}
\end{aligned}$
And
$1_{2}=\frac{a_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}=\frac{-1}{9}, m_{2}=\frac{b_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}=\frac{8}{9}, n_{2}=\frac{c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}=\frac{4}{9}$
$\therefore$ From eq.(1), we have
$\begin{array}{l}
\cos \theta=\left|\left(\frac{2}{\sqrt{38}}\right) \times\left(\frac{-1}{9}\right)+\left(\frac{5}{\sqrt{38}}\right) \times\left(\frac{8}{9}\right)+\left(\frac{-3}{\sqrt{38}}\right) \times\left(\frac{4}{9}\right)\right| \\
\quad=\left|\frac{-2+40-12}{9 \sqrt{38}}\right|=\left|\frac{40-12}{9 \sqrt{38}}\right|=\frac{26}{9 \sqrt{38}} \\
\theta=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)
\end{array}$

(ii) $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$
Here, $a_{1}=2, b_{1}=2, c_{1}=1$ and
$\mathrm{a}_{2}=4, \mathrm{~b}_{2}=1, \mathrm{c}_{2}=8$
Here, it is known to us that
$\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}=\sqrt{2^{2}+2^{2}+1^{2}}=\sqrt{4+4+1}=\sqrt{9}=3$
And
$\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}=\sqrt{4^{2}+1^{2}+8^{2}}=\sqrt{16+1+64}=\sqrt{81}=9$
From eq.(2), we have
$1_{1}=\frac{a_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\frac{2}{3}, m_{1}=\frac{b_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\frac{2}{3}, n_{1}=\frac{c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}=\frac{1}{3}$
And
$1_{2}=\frac{\mathrm{a}_{2}}{\sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}=\frac{4}{9}, \mathrm{~m}_{2}=\frac{\mathrm{b}_{2}}{\sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}=\frac{1}{9}, \mathrm{n}_{2}=\frac{\mathrm{c}_{2}}{\sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}=\frac{8}{9}$
$\therefore$ From eq. (1), we have
$\begin{array}{l}
\cos \theta=\left|\left(\frac{2}{3} \times \frac{4}{9}\right)+\left(\frac{2}{3} \times \frac{1}{9}\right)+\left(\frac{1}{3} \times \frac{8}{9}\right)\right|=\left|\frac{8+2+8}{27}\right|=\frac{18}{27}=\frac{2}{3} \\
\theta=\cos ^{-1}\left(\frac{2}{3}\right)
\end{array}$