(i) To find the work done by the lorry's driving force, we calculate the gain in potential energy (PE) and the work done against resistance.

Gain in PE = mass × gravity × height = 15,000 kg × 9.8 m/s² × 500 m × \(\sin 2.5^{\circ}\)

\(Work done against resistance = resistance × distance = 800 N × 500 m\)

\(Total work done = Gain in PE + Work done against resistance = 3,271,454 J + 400,000 J = 3,670,000 J or 3670 kJ\)

(ii) For the return journey, we use the work-energy principle.

\(Work done by driving force = 2000 N × 500 m = 1,000,000 J\)

Gain in kinetic energy (KE) = \(\frac{1}{2} \times 15,000 \times (v^2 - 20^2)\)

\(Using the equation: Gain in KE = Loss in PE - Work done against resistance + Work done by driving force\)

\(\frac{1}{2} \times 15,000 \times (v^2 - 20^2) = 3,271,454 - 400,000 + 1,000,000\)

Solving for \(v\), we find \(v = 30.3 \text{ m/s}\)