(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:

\(3000 - 2400 = 16000a\)

\(600 = 16000a\)

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

(a)(i) The power developed by the engine is given by the formula:

\(P = F \times v\)

where \(F = 650 \text{ N}\) and \(v = 30 \text{ m/s}\).

\(P = 650 \times 30 = 19500 \text{ W} = 19.5 \text{ kW}\).

(a)(ii) The increase in power is 9 kW, so the new power is:

\(19500 + 9000 = DF \times 30\)

\(DF = 1300a + 650\)

Solving for \(a\):

\(9000 = DF \times 30\)

\(DF = 1300a\)

\(a = \frac{3}{13} = 0.231 \text{ m/s}^2\)

(b) Using the power equation:

\(DF \times v = 11500\)

Using Newton's second law:

\(11500/v + 1300 \times g \times 0.08 - (1000 + 20v) = 0\)

Solving for \(v\):

\(v = 25 \text{ m/s}\)

(i) The power output is given by the formula:

\(P = F \times v\)

where \(P = 240\) W and \(v = 6\) m/s. Thus, the driving force \(F\) is:

\(F = \frac{240}{6} = 40\) N

Using Newton's Second Law:

\(F - R = ma\)

where \(m = 80\) kg and \(a = 0.3\) m/s². Therefore:

\(40 - R = 80 \times 0.3\)

\(R = 40 - 24 = 16\) N

(ii) For steady speed, the driving force equals the resistance:

\(\frac{240}{v} = 16\)

Solving for \(v\):

\(v = \frac{240}{16} = 15\) m/s

(iii) On an incline, using Newton's Second Law:

\(\frac{240}{4} - 16 - 80g \sin 3^{\circ} = 80a\)

\(60 - 16 - 80 \times 9.8 \times \sin 3^{\circ} = 80a\)

\(60 - 16 - 41.04 = 80a\)

\(2.96 = 80a\)

\(a = \frac{2.96}{80} = 0.0266\) m/s²

The power \(P\) of the crane is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity.

The force \(F\) required to lift the load is equal to the weight of the load, which is \(F = mg\), where \(m = 1250\) kg is the mass and \(g = 9.8\) m s^-2 is the acceleration due to gravity.

Thus, \(F = 1250 \times 9.8\).

Given that the power \(P = 20,000\) W (since 20 kW = 20,000 W), we have:

\(20,000 = V \times 1250 \times 9.8\).

Solving for \(V\):

\(V = \frac{20,000}{1250 \times 9.8}\).

\(V = 1.6\) m s^-1.

(i) Using Newton's Second Law, the equation of motion up the hill is:

\(DF - 1550 - 1400g \sin 4^{\circ} = 1400 \times 0.4\)

Solving for the driving force \(DF\):

\(DF = 3086.59 \ldots \text{ N}\)

Using the power formula \(P = Fv\), where \(P = 30000 \text{ W}\):

\(30000 = (1400 \times 0.4 + 1550 + 1400g \sin 4^{\circ})v\)

Solving for \(v\):

\(v = 9.72 \text{ m s}^{-1}\)

(ii) For constant speed, the net force is zero:

\(DF - 1550 - 1400g \sin 4^{\circ} = 0\)

Solving for \(DF\):

\(DF = 2526.59 \ldots \text{ N}\)

Using the power formula \(P = Fv\) with \(v = 40 \text{ m s}^{-1}\):

\(P_{\text{max}} = (1550 + 1400g \sin 4^{\circ}) \times 40\)

\(P = 101063.6 \ldots \text{ W}\)

Thus, the maximum possible power is 101 kW.