$x^2+y^2+2g_{1}x+2f_{1}y+c^{\text{'}}=0$

and $x^2+y^2+2g_{2}x+2f_{2}y+c^{\text{'}\text{'}}=0$

is given by $\cos \theta =\frac{{\left(c_1{c}_2\right)}^2-r_1^2-r_2^2}{2r_1{r}_2}$ where,

$c_1{c}_2$ is the distance between the centres of the two circles and $r_1 \& r_2$ are the radius of the circles.

Now,

$C_1\equiv x^2+y^2-12x-6y+41=0$

Here, $g_1=-6, f_1=-3, c^{\text{'}}=41$

Therefore,

$c_1\equiv \left(6,3\right)$

 $r_1=\sqrt{g_1^2+f_1^2-c^{\text{'}}}$

$=\sqrt{36+9-41}=2$

And,

$C_2\equiv x^2+y^2+kx+6y-59=0$    

Here, $g_2=\frac{k}{2}, f_2=3, c^{\text{'}\text{'}}=-59$

Therefore,

$c_2\equiv \left(-\frac{k}{2},-3\right)$

$r_2=\sqrt{\frac{k^2}{4}+9+59}$

$=\sqrt{\frac{k^2}{4}+68}$     $$

Now, distance between centres is given by

$c_1{c}_2=\sqrt{{\left(6+\frac{k}{2}\right)}^2+36}$

Now, we have $\theta =45°$. Therefore,

$\cos 45°=\frac{{\left(c_1{c}_2\right)}^2-r_1^2-r_2^2}{2r_1{r}_2}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{{\left(\sqrt{{\left(6+{\displaystyle \frac{k}{2}}\right)}^2+36}\right)}^2-4-\left({\displaystyle \frac{k^2}{4}}+68\right)}{2\times 2\times \sqrt{{\displaystyle \frac{k^2}{4}}+68}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{{\left(6+{\displaystyle \frac{k}{2}}\right)}^2+36-4-\left({\displaystyle \frac{k^2}{4}+68}\right)}{2\sqrt{{\displaystyle k^2+272}}}$

$\Rightarrow \sqrt{2}\left(\sqrt{k^2+272}\right)={\left(6+\frac{k}{2}\right)}^2+32-\left(\frac{k^2}{4}+68\right)$

$\Rightarrow \sqrt{2}\left(\sqrt{k^2+272}\right)=6k$

$\Rightarrow 2\left(k^2+272\right)=36k^2$

$\Rightarrow k^2+272=18k^2$

$\Rightarrow 17k^2=272$

$\Rightarrow k^2=16$

$\Rightarrow k=\pm 4$

Hence, the value of $k$ is $-4.$.