(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.