(a) The power exerted by the cyclist is given by \(P = Fv\), where \(F\) is the forward force and \(v\) is the velocity. Thus, \(F = \frac{150}{4} = 37.5 \text{ N}\).

Using Newton's second law, \(F - 20 = m \times 0.25\).

Substitute \(F = 37.5\):

\(37.5 - 20 = m \times 0.25\)

\(17.5 = m \times 0.25\)

\(m = \frac{17.5}{0.25} = 70 \text{ kg}\).

(b) Resolving forces up the plane, the equation is:

\(\frac{150}{3} - 20 - 70g \sin \theta = 0\).

Solving for \(\theta\):

\(50 - 20 = 70g \sin \theta\)

\(30 = 70g \sin \theta\)

\(\sin \theta = \frac{30}{70g}\)

\(\theta = \sin^{-1} \left( \frac{3}{m} \right)\)

\(\theta = 2.5^\circ\) to 1 d.p.