(i) The power supplied by the engine is 30 kW, which is 30000 W. The work done by the engine is given by:

\(\text{Work Done} = \text{Power} \times \text{Time} = 30000 \times 100 = 3000000 \text{ J}\)

The work done against the resistance is:

\(\text{Work Done} = \text{Resistance} \times \text{Distance} = 750 \times AB\)

Equating the two expressions for work done:

\(750 \times AB = 3000000\)

\(AB = \frac{3000000}{750} = 4000 \text{ m}\)

(ii) Using the equation of motion \(v^2 = u^2 + 2as\), where \(u = 40 \text{ m/s}\), \(v = 20 \text{ m/s}\), and \(a\) is the acceleration:

\(20^2 = 40^2 + 2(-1.25)BC\)

Solving for \(BC\):

\(400 = 1600 - 2.5BC\)

\(2.5BC = 1200\)

\(BC = \frac{1200}{2.5} = 480 \text{ m}\)

(iii) The work done by the engine is:

\(30000 \times 14 = 420000 \text{ J}\)

The gain in kinetic energy is:

\(\frac{1}{2} \times 600 \times (30^2 - 20^2) = 150000 \text{ J}\)

The work done against resistance is:

\(750 \times CD = 420000 - 150000\)

\(750 \times CD = 270000\)

\(CD = \frac{270000}{750} = 360 \text{ m}\)