Any tangent to parabola $y^{2}=8 x$ is $y=m x+\frac{2}{m} \ldots$(i)
It touches the circle $x^{2}+y^{2}-12 x+4=0$,
if the length of perpendicular from the centre $(6,0)$ is equal to radius $\sqrt{32}$.
$\therefore \frac{6 m +\frac{2}{ m }}{\sqrt{ m ^{2}+1}}=\pm \sqrt{32}$
$\Rightarrow \left(3 m +\frac{1}{ m }\right)^{2}=8\left( m ^{2}+1\right)$
$\Rightarrow \left(3 m ^{2}+1\right)^{2}=8\left( m ^{4}+ m ^{2}\right) $
$\Rightarrow m ^{4}-2 m ^{2}+1=0$
$ \Rightarrow m =\pm 1$
Hence, the required tangents are $y=x+2$ and $y=-x-2$.