find the slope of the line which makes an angle of 45 with the line 3x-y+5=0

let the slope of the required line be m.
the equation of the given line is 3x-y+5=0....(1)
$\Rightarrow y=3x+5$
slope of the line (1) is 3.
the angle between the lines is 45 deg
if the slope of two lines are $m_1 and m_2$ .
and the angle between them is $\theta$.
then
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
therefore
$$\begin{gathered} \tan 45=\left|\frac{3-m}{1+3m}\right| \\ \left|\frac{3-m}{1+3m}\right|=1 \\ \Rightarrow \frac{3-m}{1+3m}=\pm 1 \\ \frac{3-m}{1+3m}=1 or \frac{3-m}{1+3m}=-1 \\ 3-m=1+3m or 3-m=-(1+3m) \\ 4m=2 or 2m=-4 \\ m =\frac{1}{2} or m=-2 \end{gathered}$$

hope this helps you