Answer: \(\boxed{v_A=\dfrac{3-e}{4}u}\), \(\boxed{v_B=\dfrac34(1+e)u}\), and \(\boxed{e=\dfrac35}\).

(i) Take the original direction of motion of \(A\) as positive. Let the velocities immediately after the first collision be \(v_A\) and \(v_B\).

Conservation of momentum gives

\[3mv_A+mv_B=3mu.\]

Dividing by \(m\),

\[3v_A+v_B=3u.\]

Newton's law of restitution gives

\[v_B-v_A=e(u-0)=eu.\]

So \(v_B=v_A+eu\). Substitute this into the momentum equation:

\[3v_A+v_A+eu=3u.\]

Therefore

\[v_A=\frac{3-e}{4}u.\]

Then

\[v_B=v_A+eu=\frac{3-e}{4}u+eu=\frac34(1+e)u.\]

(ii) After \(B\) hits the wall, its velocity is reversed and multiplied by \(\frac34\). Therefore

\[v'_B=-\frac34v_B=-\frac{9}{16}(1+e)u.\]

When \(A\) and \(B\) collide again, \(A\) is brought to rest. Let the final velocity of \(B\) be \(V_B\).

Conservation of momentum for the second collision gives

\[mV_B=3mv_A+mv'_B.\]

So

\[V_B=3v_A+v'_B.\]

Substitute \(v_A=\frac{3-e}{4}u\) and \(v'_B=-\frac{9}{16}(1+e)u\):

\[V_B=3\left(\frac{3-e}{4}u\right)-\frac{9}{16}(1+e)u.\]

Hence

\[V_B=\frac{3(9-7e)}{16}u.\]

Restitution for the second collision gives

\[V_B=e(v_A-v'_B),\]

because the speed of approach is \(v_A-v'_B\). Thus

\[V_B=e\left\{\frac{3-e}{4}u+\frac{9}{16}(1+e)u\right\}.\]

So

\[V_B=\frac{e(21+5e)}{16}u.\]

Equate the two expressions for \(V_B\):

\[3(9-7e)=e(21+5e).\]

Expanding gives

\[27-21e=21e+5e^2,\]

so

\[5e^2+42e-27=0.\]

Factor or use the quadratic formula:

\[(5e-3)(e+9)=0.\]

The coefficient of restitution must be between \(0\) and \(1\), so

\[\boxed{e=\frac35}.\]