The slopes of the lines, which make an angle $45^{\circ}$ with the line $3 x-y=-5$, are

The given line is $3x - y = -5$, which can be written in the slope-intercept form $y = 3x + 5$.

The slope of this line is $m_1 = 3$.

Let the slope of the required lines be $m$. The angle between the lines is given as $45^{\circ}$.

Using the formula for the angle between two lines, $\tan \theta = \left| \dfrac{m - m_1}{1 + m m_1} \right|$, we have:

$\tan 45^{\circ} = \left| \dfrac{m - 3}{1 + 3m} \right|$

$1 = \left| \dfrac{m - 3}{1 + 3m} \right|$

This leads to two cases:

Case 1: $\dfrac{m - 3}{1 + 3m} = 1$

$m - 3 = 1 + 3m$

$-2m = 4 \Rightarrow m = -2$

Case 2: $\dfrac{m - 3}{1 + 3m} = -1$

$m - 3 = -1 - 3m$

$4m = 2 \Rightarrow m = \dfrac{1}{2}$

The slopes of the lines are $-2$ and $\dfrac{1}{2}$.

Answer: $-2, \dfrac{1}{2}$