(i) Using Newton's second law, the net force on the car is given by:

\(F = ma = 1200 \times 0.5 = 600 \text{ N}\)

The driving force (DF) minus the resistance is equal to the net force:

\(\text{DF} - 400 = 600\)

Thus, the driving force is:

\(\text{DF} = 1000 \text{ N}\)

The power is given by:

\(P = Fv\)

Substituting the values, we have:

\(20000 = 1000v\)

Solving for \(v\):

\(v = 20 \text{ m/s}\)

(ii) At maximum speed, the acceleration is zero, so the net force is zero:

\(\frac{20000}{v} - 400 = 0\)

Solving for \(v\):

\(\frac{20000}{v} = 400\)

\(v = 50 \text{ m/s}\)

(iii) The work done is given by:

\(\Delta W = 1500 \text{ kJ} = 1500000 \text{ J}\)

The distance \(d\) is:

\(d = \frac{1500000}{400} = 3750 \text{ m}\)

The time taken \(t\) is:

\(t = \frac{3750}{50} = 75 \text{ s}\)