(i) To find the work done by the driving force from A to B, we first calculate the increase in kinetic energy (KE) of the lorry:

Increase in KE = \(\frac{1}{2} \times 12500 \times (25^2 - 17^2)\)

\(= \frac{1}{2} \times 12500 \times (625 - 289)\)

\(= \frac{1}{2} \times 12500 \times 336\)

\(= 2100 \text{ kJ}\)

The total work done by the driving force is the sum of the work done against resistance and the increase in KE:

\(Work done by driving force = 2100 kJ + 5000 kJ = 7100 kJ\)

(ii) For the section from B to C, we use the work-energy principle. The work done by the driving force is 3300 kJ, and the work done against resistance is \(4800 \times 500 \div 1000 = 2400 \text{ kJ}\).

The potential energy (PE) gain is given by:

\(PE gain = (3300 + 2100) - 2400\)

\(= 3000 \text{ kJ}\)

Using \(mgh = \text{PE gain}\), we find the height \(h\):

\(12500 \times 10 \times h = 3000 \times 1000\)

\(125000h = 3000000\)

\(h = \frac{3000000}{125000} = 24 \text{ m}\)