(i) The power \(P\) of the engine is given as \(22.5 \, \text{kW} = 22500 \, \text{W}\). The force \(F\) exerted by the engine is given by \(F = \frac{P}{v}\), where \(v = 18 \, \text{m/s}\). Thus, \(F = \frac{22500}{18} = 1250 \, \text{N}\).

Using Newton's second law, the net force is \(F - R = ma\), where \(m = 600 \, \text{kg}\) and \(a = 1.4 \, \text{m/s}^2\). Therefore, \(1250 - R = 600 \times 1.4\).

Solving for \(R\), we get \(R = 1250 - 840 = 410 \, \text{N}\).

(ii) When the car is moving at a constant speed of \(15 \, \text{m/s}\), the power \(P\) is given by \(P = R \times v\). Thus, \(P = 410 \times 15 = 6150 \, \text{W}\).