(i) When the lorry is going up the hill at a constant speed, the force required is given by:

\(F = 1480 + 7850g \sin 3^{\circ}\)

Calculating the force:

\(F = 1480 + 7850 \times 9.8 \times \sin 3^{\circ} = 5588 \text{ N}\)

The power is given by:

\(P = Fv = 5588 \times 10 = 55,880 \text{ W}\)

Rounding to three significant figures, the power is 55,900 W.

(ii) When the lorry is going down the hill with acceleration, use Newton's Second Law:

\(F + 7850g \sin 3^{\circ} - 1480 = 7850 \times 0.8\)

Solving for the force:

\(F = 3652 \text{ N}\)

The power is given by:

\(P = Fv = 3652 \times 15 = 54,780 \text{ W}\)

Rounding to three significant figures, the power is 54,800 W.