(a) To find the distance \(AB\), we need to determine when the cyclist comes to rest at \(B\). The velocity \(v\) is the derivative of \(s\) with respect to \(t\):

\(v = \frac{ds}{dt} = 0.004(150t - 3t^2) = 0.6t - 0.012t^2\).

Set \(v = 0\) to find when the cyclist stops:

\(0.6t - 0.012t^2 = 0\)

\(t(0.6 - 0.012t) = 0\)

\(t = 0\) or \(t = 50\).

At \(t = 50\), substitute into \(s\):

\(s = 0.004(75 \times 50^2 - 50^3) = 0.004(187500 - 125000) = 0.004 \times 62500 = 250 \text{ m}\).

Thus, the distance \(AB = 250 \text{ m}\).

(b) The maximum velocity occurs when the acceleration is zero. Differentiate \(v\) to find acceleration \(a\):

\(a = \frac{dv}{dt} = 0.6 - 0.024t\).

Set \(a = 0\) to find the time of maximum velocity:

\(0.6 - 0.024t = 0\)

\(t = 25\).

Substitute \(t = 25\) into \(v\):

\(v = 0.004(150 \times 25 - 3 \times 25^2) = 0.004(3750 - 1875) = 0.004 \times 1875 = 7.5 \text{ ms}^{-1}\).

Thus, the maximum velocity is 7.5 \text{ ms}^{-1}.