(a) The power of the lorry's engine can be calculated using the formula:

\(P = \frac{\text{Work Done}}{\text{Time}}\)

Given that the work done is 750000 J and the time is 10 s, we have:

\(P = \frac{750000}{10} = 75000 \text{ W}\)

Therefore, the power is 75 kW.

(b) To find the acceleration, we first find the driving force (DF) using the power formula:

\(P = \text{DF} \times v\)

\(75000 = \text{DF} \times 25\)

\(\text{DF} = \frac{75000}{25} = 3000 \text{ N}\)

Using Newton's second law, \(F = ma\), where the net force \(F\) is the driving force minus the resistance force:

\(3000 - 2400 = 16000a\)

\(600 = 16000a\)

\(a = \frac{600}{16000} = 0.0375 \text{ m s}^{-2}\)