(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:

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