Shortest distance between two curves occured along the common normal.

$\therefore$ Normal to $y^2=4x$ at $\left(m^2,2m\right)$ is

$y+mx-2m-m^3=0$

Normal to $y^2=2\left(x-3\right)$ at $\left(\frac{m^2}{2}+3,m\right)$ is

$y+m\left(x-3\right)-m-\frac{m^3}{2}=0$

Both normals are same, if

$-2m-m^3=-4m-\frac{1}{2}{m}^3\Rightarrow m=0,2$

So, points will be $(4,4)$ and $(5,2)$ or $(4,-4)$ and $(5,-2)$.

Hence, shortest distance will be 

$\sqrt{(1+4)}=\sqrt{5}$ unit.