The image of the point $\left(4,-13\right)$  with respect to the line $2x+y+6=0$  is

$\frac{h-4}{2}=\frac{k+13}{1}=\frac{-2\left(2\times 4+1\left(-13\right)+6\right)}{2^2+1^2}$

$\frac{h-4}{2}=\frac{k+13}{1}=\frac{-2\left(8-13+6\right)}{5}$

$\frac{h-4}{2}=\frac{k+13}{1}=-2\left(\frac{1}{5}\right)$

$\frac{h-4}{2}=\frac{-2}{5}$

$5h-20=-4$

$5h=-4+20=16$

$h=\frac{16}{5}$

$k+13=\frac{-2}{5}$

$k=\frac{-2}{5}-13$

$=\frac{-2-65}{2}$

$=-\frac{67}{2}$

$\left(h,k\right)=\left(\frac{16}{5},\frac{-67}{2}\right)$