(i) The work done against resistance is \(800 \times 500 = 400,000 \text{ J}\).

The total work done by the driving force is \(2,800,000 \text{ J}\).

Using the equation: \(\text{Work done by driving force} = \text{Potential energy gain} + \text{Work done against resistance}\)

\(2,800,000 = \text{PE gain} + 400,000\)

\(\text{PE gain} = 2,400,000 \text{ J}\)

\(\text{PE gain} = mgh = 16,000 \times 9.8 \times 500 \sin \alpha\)

\(2,400,000 = 16,000 \times 9.8 \times 500 \sin \alpha\)

Solving for \(\alpha\), \(\alpha = 1.7^\circ\).

(ii) The kinetic energy gain is given by:

\(\text{KE gain} = \text{Work done by driving force} + \text{PE loss} - \text{Work done against resistance}\)

\(\text{KE gain} = 2,400,000 + 2,400,000 - 800,000 = 4,000,000 \text{ J}\)

Using the kinetic energy formula: \(\frac{1}{2} m (v^2 - 20^2) = 4,000,000\)

\(\frac{1}{2} \times 16,000 \times (v^2 - 400) = 4,000,000\)

\(8,000 \times (v^2 - 400) = 4,000,000\)

\(v^2 - 400 = 500\)

\(v^2 = 900\)

\(v = 30 \text{ m s}^{-1}\)