(i) Using the work-energy principle, the work done by the pulling force is equal to the increase in kinetic energy minus the decrease in potential energy plus the work done against resistance.

Given: Work done by pulling force = 1150 J, mass \(m = 16\) kg, distance \(s = 50\) m, \(\sin \theta = 0.05\), final speed \(v = 10\) m/s.

Potential energy change \(= mgh = 16 \times g \times 50 \times 0.05\).

Kinetic energy change \(= \frac{1}{2} m v^2 = \frac{1}{2} \times 16 \times 10^2\).

1150 = \(\frac{1}{2} \times 16 \times 10^2 - 16 \times g \times 50 \times 0.05 + \text{WD by resistance}\).

Solving for work done by resistance, \(\text{WD by resistance} = 750\) J.

(ii) When the block is pulled up from B to A, the work done by the pulling force is equal to the gain in kinetic energy plus the gain in potential energy plus the work done against resistance.

1150 = increase in kinetic energy + \(16 \times g \times 50 \times 0.05 + 750\).

Since the work done by the pulling force and the work done against resistance are the same as in the downward motion, the gain in kinetic energy is zero.

Thus, the speed at A is the same as the speed at B.