(a) At the end of the first stage, the speed is \(0+0.5(10)=5\text{ m s}^{-1}\). For the second stage use \(v^2=u^2+2as\):

\[v^2=5^2+2(1.5)(13)=64\quad\Longrightarrow\quad\boxed{v=8}.\]

(b)(i) The first-stage distance is \(\frac12(0.5)(10)^2=25\text{ m}\); the constant-speed stage covers \(8(5)=40\text{ m}\). The total distance is

\[25+13+40+28=\boxed{106\text{ m}}.\]

(b)(ii) The second stage takes \((8-5)/1.5=2\text{ s}\). During the final stage the mean speed is \((8+0)/2=4\text{ m s}^{-1}\), so the time is \(28/4=7\text{ s}\).

\[\text{Average speed}=\frac{106}{10+2+5+7}=\frac{53}{12}\text{ m s}^{-1}=\boxed{4.42\text{ m s}^{-1}}\quad\text{(3 s.f.)}.\]