Solution

Since the shortest distance between the two curves happens to be at the normal which is common to both the cuves.
Therefore
The normal to the curve $$\displaystyle y^2=4x $$ at $$\displaystyle (m^2,2m) $$ is given by:
$$\displaystyle (y- 2m)=- m (x-m^2) $$
i.e., $$\displaystyle y-2m= -mx+m^3 $$
i.e., $$\displaystyle y-2m+mx-m^3 = 0 $$
i.e., $$\displaystyle y+mx-2m-m^3 = 0 $$

And the normal to the curve $$\displaystyle y^2=2x-6 $$ at $$\displaystyle \left(\dfrac{1}{2}m^2+3,m\right) $$ is given by:
$$\displaystyle (y-m) = -m \left( x - \dfrac{1}{2}m^2-3 \right) $$
i.e., $$\displaystyle y-m = -mx + \dfrac{1}{2}m^3+3m $$
i.e., $$\displaystyle y-m +mx - \dfrac{1}{2}m^3-3m =0$$
i.e., $$\displaystyle y+mx -4m- \dfrac{1}{2}m^3=0$$

These two normals are common if:
$$\displaystyle y+mx-2m-m^3 = $$ $$\displaystyle y+mx -4m- \dfrac{1}{2}m^3$$
i.e., $$\displaystyle -2m-m^3 = $$ $$\displaystyle -4m- \dfrac{1}{2}m^3$$
i.e., $$\displaystyle m^3-4m =0 $$
i.e., $$\displaystyle m(m^2-4) = $$
i.e., $$\displaystyle m(m+2)(m-2) = $$
Therefore, $$\displaystyle m = 0,2,-2 $$

Thus, the points are: $$\displaystyle (4,4) $$ and $$\displaystyle (5,2) $$

And the distance is: $$\displaystyle d = \sqrt{(5-4)^2+ (2-4)^2} $$
i.e., $$\displaystyle d = \sqrt{1+ 4} $$
i.e., $$\displaystyle d = \sqrt{5} $$