Given circles,
$$
\begin{aligned}
x^2+y^2-12 x-6 y+41 & =0 \\
\text { and } x^2+y^2+k x+6 y-59 & =0
\end{aligned}
$$
Centre of first circle is $(6,3)$ and radius
$$
r_1=\sqrt{36-9-4}=2
$$
Centre of second circle is $\left(\frac{k}{2},-3\right)$ and radius
$$
r=\sqrt{\frac{k^2}{4}+9+59}=\sqrt{\frac{k^2}{4}+68}
$$
Now, distance between centres is
$$
\begin{aligned}
& d=\sqrt{\left(\frac{k}{2}-6\right)^2+6^2} \\
& d=\sqrt{\frac{k^2}{4}-6 k+72}
\end{aligned}
$$
Now, $\cos \theta=\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}$
$$
\begin{aligned}
& \Rightarrow \quad \cos 45^{\circ}=\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2} \\
& \Rightarrow 2 \cdot \sqrt{2} \sqrt{\frac{k^2}{4}+68}=4+\frac{k^2}{4}+68-\frac{k^2}{4}+6 k-72 \\
& \Rightarrow \quad 2 \sqrt{2} \sqrt{\frac{k^2}{4}+68}=6 k \\
& \Rightarrow \quad \frac{k^2}{4}+68=\frac{36 k^2}{8} \Rightarrow \frac{17 k^2}{4}=68
\end{aligned}
$$
$\begin{aligned} & \Rightarrow \quad k^2=16 \Rightarrow k^2= \pm 4 \\ & \text { Hence, } \quad k=-4 \\ & \end{aligned}$