Given, $y^{2}=8 x$
$\therefore \,\,\,4 a=8 \Rightarrow a=2$
Any tangent of parabola is,
$y=m x+\frac{a}{m}$
$ \Rightarrow m x-y+\frac{2}{m}=0$
If it is a tangent to the circle $x^{2}+y^{2}=2,$ then perpendicular from centre (0,0) is equal to radius $\sqrt{2}$
$\therefore \frac{\frac{2}{m}}{\sqrt{m^{2}+1}}=\sqrt{2}$
$\Rightarrow \,\,\, \frac{4}{m^{2}}=2\left(m^{2}+1\right)$
$\Rightarrow \,\,\, m^{4}+m^{2}-2=0$
$\Rightarrow \,\,\,\left(m^{2}+2\right)\left(m^{2}-1\right)=0$
$\Rightarrow \,\,\, m=\pm 1$
Hence, the common tangents are $y=\pm(x+2)$