(i) The power supplied by the engines is 4080 kW, which is 4080000 W. The speed of the train is 85 m s^-1. Using the formula for power \(P = Fv\), where \(F\) is the force and \(v\) is the velocity, we have:

\(F = \frac{P}{v} = \frac{4080000}{85} = 48000 \text{ N}\)

Thus, the resistance force is 48,000 N.

(ii) On the incline, the driving force \(DF\) is given by:

\(DF = \frac{P}{v}\)

Using Newton's second law, the net force is zero for constant speed, so:

\(DF - 48000 - 490000 \times g \times \frac{1}{200} = 0\)

Solving for \(DF\):

\(DF = 48000 + 490000 \times 9.8 \times \frac{1}{200}\)

\(DF = 48000 + 24010 = 72010 \text{ N}\)

The power required is:

\(P = DF \times v = 72010 \times 85 = 6120850 \text{ W} = 6.16 \text{ MW}\)

(i) The power \(P\) is given by \(P = Fv\), where \(F\) is the resistive force. At the greatest speed, \(v = 56 \text{ m s}^{-1}\), the resistive force \(F = 40 \times 56\). Therefore, \(P = 40 \times 56 \times 56 = 125440 \text{ W} = 125 \text{ kW}\).

(ii) The driving force \(F_d\) is \(\frac{125440}{32}\). Using Newton's second law, \(F_d - 40 \times 32 = 1400a\). Solving for \(a\), we get \(a = 1.89 \text{ m s}^{-2}\).

(iii) For the car on an incline, the equation is \(\frac{60000}{50} + 1400g \sin \theta - 40 \times 50 = 0\). Solving for \(\sin \theta\), we have \(\sin \theta = \frac{800}{14000}\). Thus, \(\theta = 3.3^\circ\).

(a) To find the power of the lorry's engine, use the formula for power: \(P = Fv\). At point A, \(P = 20F\) and at point B, \(P = 25F\). Given the resistance force is 6000 N, the net force at A is \(F - 6000\) and at B is \(F - 6000\).

Using Newton's second law, at A: \(\frac{P}{20} - 6000 = 15000a\) and at B: \(\frac{P}{25} - 6000 = 15000 \times 0.5a\).

Solving these equations simultaneously, we find \(P = 200,000 \text{ W} = 200 \text{ kW}\) and \(a = \frac{4}{15} \text{ m/s}^2\).

(b) For the lorry ascending the hill, the equation becomes \(\frac{200,000}{v} - 6000 - 15000g \sin(1^{\circ}) = 0\). Solving for \(v\), we find the steady speed \(v = 23.2 \text{ m/s}\).

To find the power of the engine, we first need to determine the driving force (DF) using the equation for power:

\(DF = \frac{P}{v}\)

where \(P\) is the power and \(v = 15 \text{ m/s}\) is the speed of the train.

Using Newton's second law, the equation for the forces acting on the train is:

\(DF - 240,000g \sin 4^{\circ} - 18,000 = 240,000 \times (-0.2)\)

Solving for \(DF\):

\(DF = 240,000 \times 9.8 \times \sin 4^{\circ} + 18,000 - 240,000 \times 0.2\)

Substitute \(DF\) back into the power equation:

\(P = DF \times 15\)

Calculate \(P\) to find:

\(P = 2,060,000 \text{ W}\)

(i) The driving force when the car is at its greatest speed is given by the resistive force: \(35 \times 60\). The power is then \(35 \times 60^2 = 126000 \text{ W}\).

(ii) The driving force \(\text{DF}\) is \(\frac{126000}{30}\). Using Newton's second law: \(\text{DF} - 35 \times 30 = 1200a\). Solving for \(a\), we have \(a = \frac{3150}{1200} = \frac{21}{8} = 2.625 \text{ m s}^{-2}\).

(iii) The driving force \(\text{DF} = \frac{126000}{v}\). The equation for forces is \(\frac{126000}{v} = 35v + 1200g \times \frac{7}{48}\). Simplifying gives \(35v^2 + 1750v - 126000 = 0\) or \(v^2 + 50v - 3600 = 0\). Solving this quadratic equation gives \(v = 40 \text{ m s}^{-1}\).