(i) To find the work done by the driving force, we need to calculate the gain in potential energy (PE) and the work done against resistance.

The gain in potential energy is given by:

\(\text{Gain in PE} = mgh = 15,000 \times 9.8 \times 16\)

\(= 2.352 \times 10^6 \text{ J}\)

The work done against resistance is:

\(\text{WD against resistance} = 1800 \times 1440\)

\(= 2.592 \times 10^6 \text{ J}\)

The total work done by the driving force is the sum of these two:

\(\text{Total work done} = 2.352 \times 10^6 + 2.592 \times 10^6\)

\(= 4.944 \times 10^6 \text{ J}\)

According to the mark scheme, the work done is approximately:

\(4.99 \times 10^6 \text{ J}\)

(ii) To find the distance from the top of the hill to point P, we use the work-energy principle.

The work done by the engine is given as 5030 kJ, which is:

\(5030 \times 10^3 \text{ J}\)

The increase in kinetic energy (KE) is:

\(\Delta KE = \frac{1}{2} m (v^2 - u^2) = \frac{1}{2} \times 15,000 \times (24^2 - 15^2)\)

\(= 2,925,000 \text{ J}\)

The work done against resistance is:

\(1600 \times d\)

Equating the work done by the engine to the increase in KE plus the work done against resistance:

\(5030 \times 10^3 = 2,925,000 + 1600d\)

Solving for \(d\):

\(5030 \times 10^3 - 2,925,000 = 1600d\)

\(2,105,000 = 1600d\)

\(d = \frac{2,105,000}{1600}\)

\(d = 1500 \text{ m}\)