(a)(i) The total resistance force is the sum of the resistances on the car and caravan:

\(F = 400 + 250 = 650 \text{ N}\)

The power \(P\) is given by \(P = Fv\), where \(v = 25 \text{ m s}^{-1}\). Thus,

\(P = 650 \times 25 = 16250 \text{ W} = 16.25 \text{ kW}\)

(a)(ii) The engine's power is increased to 39 kW, so the driving force \(DF\) is:

\(DF = \frac{39000}{25} = 1560 \text{ N}\)

Using Newton's second law for the system:

\(1560 - 650 = 2400a\)

Solving for \(a\):

\(a = \frac{1560 - 650}{2400} = 0.379 \text{ m s}^{-2}\)

For the tension \(T\) in the tow-bar:

\(T - 250 = 800a\)

Substituting \(a\):

\(T = 250 + 800 \times 0.379 = 553 \text{ N}\)

(b) The car and caravan travel up a hill inclined at \(\sin^{-1}(0.05)\). The driving force is:

\(DF = 650 + 2400 \times 10 \times 0.05\)

Using \(P = Fv\):

\(32500 = (650 + 2400 \times 0.05)v\)

Solving for \(v\):

\(v = \frac{32500}{650 + 120} = 17.6 \text{ m s}^{-1}\)