Answer
$a.\quad\mathrm{F}_{\mathrm{n}\mathrm{e}\mathrm{t}}=0.$
$b.\quad\mathrm{A}_{\mathrm{c}\mathrm{m}}=0$
Work Step by Step
Before the string is cut, the masses are stationary. Net force is zero. The weight of the blocks, acting downward with magnitude $2\mathrm{m}\mathrm{g}$ equals in magnitude $\mathrm{F}$, the force of the spring that acts upwards.
$\mathrm{F}=2\mathrm{m}\mathrm{g}.$
a.
Taking up being the +y direction, when the string is cut, the lower weight experiences a force of $-\mathrm{m}\mathrm{g},$ and the upper weight experiences $\mathrm{F}-\mathrm{m}\mathrm{g}=2\mathrm{m}\mathrm{g}-\mathrm{m}\mathrm{g}=\mathrm{m}\mathrm{g}$
The net external force acting on the system is
$\mathrm{F}_{\mathrm{n}\mathrm{e}\mathrm{t}}=-\mathrm{m}\mathrm{g}+\mathrm{m}\mathrm{g}=0.$
b.
Since the net force acting on the center of mass is zero,
$\mathrm{A}_{\mathrm{c}\mathrm{m}}=0$
