(a) Take the original direction of \(A\) as positive. Conservation of momentum gives \[m(2m)-2(m+2)=2m+2(3.25).\] Rearranging, \[2m^2-4m-10.5=0\quad\Rightarrow\quad m=3.5\text{ or }-1.5.\] Mass is positive, so \(\boxed{m=3.5}\).

(b) Write the post-collision velocities as \(v_B,v_C\) in the same positive direction. Conservation of momentum gives \[2(3.25)=2v_B+4v_C\quad\Rightarrow\quad v_B+2v_C=3.25.\] The kinetic-energy loss gives \[\frac12(2)(3.25)^2-\frac12(2)v_B^2-\frac12(4)v_C^2=6.875,\] so \(v_B^2+2v_C^2=3.6875\).

Eliminating \(v_B\), \[(3.25-2v_C)^2+2v_C^2=3.6875\quad\Rightarrow\quad6v_C^2-13v_C+6.875=0.\] The possible pairs are \[(v_B,v_C)=\left(\frac34,\frac54\right)\quad\text{or}\quad\left(\frac{17}{12},\frac{11}{12}\right).\]

Immediately after impact, \(B\) is behind \(C\); they must separate, so \(v_B\le v_C\). This excludes the second pair. Hence \(v_B=0.75\text{ m s}^{-1}\). Sphere \(A\) is behind \(B\) and still travels at \(2\text{ m s}^{-1}>0.75\text{ m s}^{-1}\), so \(\boxed{\text{there is a further collision between }A\text{ and }B}\).