(i) The power \(P\) required to overcome the resistance is given by \(P = F \times v\), where \(F = 1350\) N and \(v = 32\) m/s.

\(P = 1350 \times 32 = 43200 \text{ W} = 43.2 \text{ kW}\).

(ii) When the car travels up the hill, the driving force \(DF\) must overcome both the resistance and the component of the weight down the slope.

The component of the weight down the slope is \(1200g \sin \theta\), where \(g = 9.8 \text{ m/s}^2\).

\(DF - 1350 - 1200 \times 9.8 \times 0.1 = 0\)

\(DF = 1350 + 1200 \times 9.8 \times 0.1 = 2550 \text{ N}\)

The power \(P\) is given by \(P = DF \times v\).

\(76500 = 2550 \times v\)

\(v = \frac{76500}{2550} = 30 \text{ m/s}\)