(a)(i) The work done against the resisting force is calculated using the formula:

\(\text{Work Done} = \text{Force} \times \text{Distance}\)

The distance covered in 8 seconds at 36 m/s is:

\(36 \times 8 = 288 \text{ m}\)

Thus, the work done is:

\(1250 \times 288 = 360000 \text{ J} = 360 \text{ kJ}\)

(a)(ii) The power developed by the engine is:

\(\text{Power} = \frac{\text{Work Done}}{\text{Time}} = \frac{360000}{8} = 45000 \text{ W} = 45 \text{ kW}\)

(a)(iii) With the power increased by 12 kW, the new power is:

\(57000 \text{ W}\)

Using the formula for power:

\(P = DF \times v\)

We find the driving force (DF):

\(DF = \frac{57000}{36} = 1583.3 \text{ N}\)

Using Newton's second law:

\(1583.3 - 1250 = 1400a\)

Solving for acceleration:

\(a = \frac{333.3}{1400} = 0.238 \text{ m s}^{-2}\)

(b) For the inclined section, using the power equation:

\(64000 = 1250 \times 32 + 1400g \sin \theta \times 32\)

Solving for \(\theta\):

\(64000 = 40000 + 44800 \sin \theta\)

\(24000 = 44800 \sin \theta\)

\(\sin \theta = \frac{24000}{44800} = 0.5357\)

\(\theta = \arcsin(0.5357) \approx 3.1^\circ\)