(i) The driving force \(F\) is given by \(F = \frac{P}{v}\), where \(P = 6000 \text{ W}\) and \(v = 20 \text{ m s}^{-1}\). Thus, \(F = \frac{6000}{20} = 300 \text{ N}\).

The net force on the system is zero since the speed is constant: \(300 - R - 80 = 0\).

Solving for \(R\), we get \(R = 220 \text{ N}\).

(ii) The new driving force \(DF\) is \(\frac{12500}{25} = 500 \text{ N}\).

For the car: \(DF - T - R = 1500a\).

For the trailer: \(T - 80 = 300a\).

For the system: \(DF - 80 - R = 1800a\).

Using \(DF = 500 \text{ N}\) and \(R = 220 \text{ N}\), solve the equations:

1. \(500 - T - 220 = 1500a\)

2. \(T - 80 = 300a\)

Solving these simultaneously gives \(a = 0.111 \text{ m s}^{-2}\) and \(T = 113 \text{ N}\).