The equations of common tangents to the parabola $y^2=16x$ and the circle  $x^2+y^2=8$ are

$y=mx+c$ is a tangent to the parabola $y^2=16x$ if $c=\frac{4}{m}$ and will be a tangent to the curve $x^2+4y^2=8$ if the roots of the equation $x^2+4(mx+c{)}^2=8$ are equal.

$\begin{array}{l}\Rightarrow (8mc{)}^2=4\left(1+4m^2\right)\left(4c^2-8\right)\\ \Rightarrow c^2=8m^2+2.\end{array}$

So $c=\frac{4}{m}\Rightarrow \frac{16}{m^2}=8m^2+2.$
$\begin{array}{l}\Rightarrow 8m^4+2m^2-16=0\end{array}$

and the required tangents are  $y=x+4,y=-x-4$