(i) Using the formula for power \(P = Fv\), where \(F\) is the force and \(v\) is the velocity. The force \(F\) can be found using Newton's second law \(F = ma\), where \(m = 80\) kg and \(a = 1.2\) m s\(^{-2}\).

Thus, \(F = 80 \times 1.2 = 96\) N.

Substitute into the power formula: \(P = 96 \times 5 = 480\) W.

(ii) The power \(P = 450\) W, speed \(v = 3.6\) m s\(^{-1}\), and \(\sin \alpha = 0.035\).

Using the formula \(\frac{P}{v} - mg \sin \alpha = ma\), where \(m = 80\) kg and \(g = 9.8\) m s\(^{-2}\).

\(\frac{450}{3.6} - 80 \times 9.8 \times 0.035 = 80a\)

\(125 - 27.44 = 80a\)

\(97.56 = 80a\)

\(a = \frac{97.56}{80} = 1.21\) m s\(^{-2}\).

Using the formula for power, \(P = Fv\), where \(F\) is the force and \(v\) is the velocity, we apply Newton's second law for both uphill and downhill motion.

For uphill motion:

\(\frac{P}{3} - R - 84g \times 0.1 = 84 \times 1.25\)

For downhill motion:

\(\frac{P}{10} - R + 84g \times 0.1 = 84 \times 1.25\)

Solving these equations simultaneously:

\(P \left( \frac{1}{3} - \frac{1}{10} \right) - 168 = 0\)

\(P \times \frac{7}{30} = 168\)

\(P = 720\)

Substitute \(P = 720\) into one of the equations to find \(R\):

\(\frac{720}{3} - R - 84 \times 0.1 = 84 \times 1.25\)

\(240 - R - 8.4 = 105\)

\(R = 51\)

(i) The driving force \(DF\) is given by \(DF = \frac{P}{18}\). Using the equation \(DF - R = ma\), where \(R = 800\) N and \(a = 0.5\) m s^-2, we have:

\(\frac{P}{18} - 800 = 1400 \times 0.5\)

\(\frac{P}{18} = 800 + 700\)

\(\frac{P}{18} = 1500\)

\(P = 27000\) W

(ii) At the new speed of 25 m s^-1, the driving force \(DF\) is \(\frac{P}{25}\). Using \(DF - R = ma\), we have:

\(\frac{27000}{25} - 800 = 1400a\)

\(1080 - 800 = 1400a\)

\(280 = 1400a\)

\(a = 0.2\) m s^-2

(i) Let the speed of the train at B be \(v_B\). The power at B is 1.2 times the power at A, and the driving force at B is 0.96 times the driving force at A. Using the formula for power \(P = Fv\), we have:

\(v_B = \frac{1.2 \times 28}{0.96}\)

Calculating this gives \(v_B = 35\) m/s.

(ii) The increase in kinetic energy (KE) from A to B is given by:

\(\text{KE increase} = \frac{1}{2} \times 200,000 \times (35^2 - 28^2)\)

\(= 100,000 \times (35^2 - 28^2)\)

\(= 100,000 \times (1225 - 784)\)

\(= 100,000 \times 441\)

\(= 44.1 \times 10^6 \text{ J}\)

The work done by the engine is the sum of the increase in kinetic energy and the work done against resistance:

\(\text{WD by engine} = 44.1 \times 10^6 + 2.3 \times 10^6\)

\(= 46.4 \times 10^6 \text{ J}\)

Thus, the work done by the engine is 46,400,000 J.

First, calculate the driving force (DF) using the power formula:

\(Power = Force × Velocity\)

\(Given power = 22.5 kW = 22500 W and velocity = 18 m/s,\)

DF = \(\frac{22500}{18} = 1250\) N

Using Newton's second law, the net force is given by:

\(Net Force = Mass × Acceleration\)

Let the resistance force be \(R\). Then,

\(DF - R = 800 × 1.2\)

\(1250 - R = 960\)

\(R = 1250 - 960 = 290 \text{ N}\)