(i) The potential energy (PE) gain is calculated as:

\(\text{PE gain} = 1250 \times 10 \times 400 \times 0.125 = 625000 \text{ J}\)

The work done against resistance is:

\(\text{WD against resistance} = 800 \times 400 = 320000 \text{ J}\)

The total work done by the car's engine is the sum of the PE gain and the work done against resistance:

\(\text{WD by car's engine} = 625000 + 320000 = 945000 \text{ J} = 945 \text{ kJ}\)

(ii) Using the power and force relationship \(P = Fv\), we have:

\(\frac{v_2}{v_1} = \frac{P_2}{P_1} \times \frac{F_1}{F_2}\)

Given \(v_1 = 6 \text{ m/s}\), \(\frac{P_2}{P_1} = 5\), \(\frac{F_1}{F_2} = \frac{1}{3}\), we find \(v_2 = 10 \text{ m/s}\).

The kinetic energy (KE) gain is:

\(\text{KE gain} = \frac{1}{2} \times 1250 \times (10^2 - 6^2) = 40000 \text{ J}\)

The total work done by the car's engine is:

\(\text{WD by car's engine} = 945000 + 40000 = 985000 \text{ J} = 985 \text{ kJ}\)