(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the force and \(v\) is the speed. Given \(P = 20000\) W and \(F = 650\) N, we have:

\(20000 = 650v\)

Solving for \(v\), we get:

\(v = \frac{20000}{650} = 30.8 \text{ m s}^{-1}\)

(ii) The force required to move up the hill is the sum of the resistive force and the component of the gravitational force along the hill. This is given by:

\(F = 650 + 1400g \times \frac{1}{7}\)

Where \(g = 9.8 \text{ m s}^{-2}\). The power \(P\) is then:

\(\frac{P}{10} = 650 + 1400 \times 9.8 \times \frac{1}{7}\)

Solving for \(P\), we find:

\(P = 26500 \text{ W}\)

(iii) The power when descending is 80% of the power found in part (ii):

\(P = 0.8 \times 26500 = 21200 \text{ W}\)

Using \(P = Fv\), where \(v = 20 \text{ m s}^{-1}\), we have:

\(21200 = F \times 20\)

\(F = 1060 \text{ N}\)

The net force is given by:

\(1060 - 650 - 1400 \times 9.8 \times \frac{1}{7} = 1400a\)

Solving for \(a\), we get:

\(a = 1.72 \text{ m s}^{-2}\)