(a) The power produced by the athlete is 60 W, and he runs at a constant speed of 3 m s^-1. The resistive force can be found using the formula for power:

\(P = F imes v\)

where \(P = 60\) W and \(v = 3\) m s^-1. Solving for \(F\):

\(F = \frac{60}{3} = 20\) N

(b) The athlete runs up a 150 m inclined section with an incline of 0.8°. The initial speed is 3 m s^-1, and the driving force is 24 N. The resistive force is 13 N. Using the work-energy principle:

Potential energy change: \(PE = 84g \times 150 \times \sin(0.8^{\circ})\)

Kinetic energy change: \(KE = \frac{1}{2} \times 84 \times v^2 - \frac{1}{2} \times 84 \times 3^2\)

Work done: \(W = (24 \times 150) - (13 \times 150)\)

Setting up the work-energy equation:

\(84g \times 150 \times \sin(0.8^{\circ}) + \frac{1}{2} \times 84 \times v^2 - \frac{1}{2} \times 84 \times 3^2 = 24 \times 150 - 13 \times 150\)

Solving for \(v\):

\(1759.23 + 42v^2 - 378 = 3600 - 1950\)

\(42v^2 = 268.765\)

\(v = 2.53\) m s^-1