Answer: The speed of \(B\) after hitting the wall is \(\boxed{\dfrac{5u}{18}}\), and the second collision with \(A\) occurs \(\boxed{\dfrac12d}\) from the wall.

(i) Let the speeds of \(A\) and \(B\) immediately after their first collision be \(v_A\) and \(v_B\), in the original direction of motion of \(A\).

Conservation of momentum gives

\[mv_A+mv_B=mu.\]

Dividing by \(m\),

\[v_A+v_B=u.\]

Newton's law of restitution gives

\[v_B-v_A=\frac23u.\]

Adding the two equations gives

\[2v_B=\frac53u,\]

so

\[v_B=\frac56u.\]

When \(B\) hits the fixed wall, its speed after impact is multiplied by the coefficient of restitution \(\frac13\). Hence

\[\text{speed after hitting wall}=\frac13\cdot\frac56u=\boxed{\frac{5u}{18}}.\]

(ii) From \(v_A+v_B=u\),

\[v_A=u-\frac56u=\frac16u.\]

The time taken by \(B\) to reach the wall is

\[t_1=\frac{d}{5u/6}=\frac{6d}{5u}.\]

In this time, \(A\) moves

\[\frac{u}{6}\cdot\frac{6d}{5u}=\frac15d.\]

So, when \(B\) leaves the wall, the distance between \(A\) and the wall is

\[d-\frac15d=\frac45d.\]

After the rebound, \(A\) and \(B\) move towards each other. Their closing speed is

\[\frac16u+\frac{5}{18}u=\frac49u.\]

The time from the rebound to the second collision is therefore

\[t_2=\frac{(4/5)d}{(4/9)u}=\frac{9d}{5u}.\]

During this time, \(B\) moves away from the wall a distance

\[\frac{5u}{18}\cdot\frac{9d}{5u}=\frac12d.\]

Hence the second collision occurs \(\boxed{\frac12d}\) from the wall.