Answer
$$\theta\approx80.8^\circ$$
Work Step by Step
$$(l_1):5x-2y+4=0\hspace{2cm}(l_2):3x+5y=6$$
$$(l_1):2y=5x+4\hspace{2cm}(l_2):5y=-3x+6$$
$$(l_1):y=\frac{5}{2}x+2\hspace{2cm}(l_2):y=-\frac{3}{5}x+\frac{6}{5}$$
Let's call the angle between the pair of lines $\theta$.
We need to apply the formula built in Exercise 74
$$\tan\theta=\frac{m_2-m_1}{1+m_1m_2}$$
in which $m_1$ is the slope of $l_1$ and $m_2$ is the slope of $l_2$.
However, since $\tan\theta\gt0$, so $$\tan\theta=\Big|\frac{m_2-m_1}{1+m_1m_2}\Big|$$
The slope of line $(l_1)$: $m_1=\frac{5}{2}$
The slope of line $(l_2)$: $m_2=-\frac{3}{5}$
Therefore,
$$\tan\theta=\Big|\frac{-\frac{3}{5}-\frac{5}{2}}{1+(-\frac{3}{5})(\frac{5}{2})}\Big|$$
$$\tan\theta=\Big|\frac{-\frac{31}{10}}{1-\frac{15}{10}}\Big|$$
$$\tan\theta=\Big|\frac{-\frac{31}{10}}{-\frac{5}{10}}\Big|$$
$$\tan\theta=\Big|\frac{31}{5}\Big|=\frac{31}{5}$$
So, $$\theta\approx80.8^\circ$$