Uhmm, let a0 be the acceleration of the movable pulley with respect to the ground.
Positive direction downward.
Let a1 be the acceleration of the m1 with respect to the ground.
Let a'2 be the acceleration of the m2 with respect to the movable pulley (that's how the m2 moves in the movable pulley's view).
Let a'3 be the acceleration of the m3 with respect to the movable pulley (that's how the m3 moves in the movable pulley's view.).
So the acceleration of m2 with respect to the ground is a2 = a'2 + a0, the acceleration of m3 with respect to the ground is a3 = a'3 + a0.
And you have a0 = -a1 (The acceleration of the movable pulley and the m1 is the same in magnitude but in opposite direction).
Since the mass of the movable pulley is negligible, the resultant force acts on it must be [itex]\vec{0}[/itex].
You can use Newton's 2nd law to solve this problem.
Viet Dao,