(i) The gain in potential energy (PE) is given by the formula:

\(\Delta PE = mgh\)

where \(m = 160 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 20 \text{ m}\).

\(\Delta PE = 160 \times 9.8 \times 20 = 32,000 \text{ J}\)

(ii) The gain in kinetic energy (KE) is given by the formula:

\(\Delta KE = \frac{1}{2} mv^2\)

where \(m = 160 \text{ kg}\) and \(v = 1.25 \text{ m/s}\).

\(\Delta KE = \frac{1}{2} \times 160 \times (1.25)^2 = 125 \text{ J}\)

(iii) The total work done (WD) by the drum is the sum of the gain in potential energy, gain in kinetic energy, and the work done against resistance:

\(WD = 32,000 + 125 + 20,000 = 52,125 \text{ J}\)

The power output \(P\) is given by:

\(P = \frac{\Delta (WD)}{\Delta T}\)

where \(\Delta T = 41.7 \text{ s}\).

\(P = \frac{52,125}{41.7} = 1250 \text{ W}\)