(a) Let \(a\) be the upward acceleration of \(A\), also the downward acceleration of \(B+C\). For \(A\), \[37.5-3(10)=3a\quad\Rightarrow\quad a=2.5\text{ m s}^{-2}.\] For \(B+C\), \[(1.5+m)10-37.5=(1.5+m)2.5.\] Thus \(7.5(1.5+m)=37.5\), so \(\boxed{m=3.5}\).

(b) Before detachment, \[v^2=2(2.5)(0.8)=4\quad\Rightarrow\quad v=2\text{ m s}^{-1},\qquad t_1=\frac2{2.5}=0.8\text{ s}.\]

After detachment, take the motion of \(A\) upwards as positive. The remaining total mass is 4.5 kg, and gravity provides resultant force \(15-30=-15\text{ N}\) in this direction: \[a_2=\frac{15-30}{3+1.5}=-\frac{10}{3}\text{ m s}^{-2}.\]

The additional time before \(A\) stops is found from \(0=2-(10/3)t_2\), giving \(t_2=0.6\text{ s}\). Therefore \[t_1+t_2=0.8+0.6=\boxed{1.4\text{ s}}.\]