(i) The power of the car's engine is given by the formula:

\(\text{Power} = \text{Force} \times \text{Velocity}\)

Substituting the given values:

\(\text{Power} = 350 \times 15 = 5250 \text{ W}\)

(ii) When the car descends the hill, the forces acting along the slope are:

\(\text{Driving Force (DF)} + 1200g \sin 1^{\circ} - 350 = 1200 \times 0.12\)

Solving for DF:

\(\text{DF} = 1200 \times 0.12 - 1200g \sin 1^{\circ} + 350\)

The power is then:

\(P = \text{DF} \times 15 = 4270 \text{ W}\)

(iii) Using Newton's second law down the slope:

\(1200g \sin 1^{\circ} - 350 = 1200a\)

Using the constant acceleration formula:

\(18^2 = 20^2 + 2as\)

Solving for \(s\):

\(s = 324 \text{ m}\)

Alternatively, using energy methods:

Potential energy loss: \(1200g \times s \times \sin 1^{\circ}\)

Kinetic energy loss: \(\frac{1}{2} \times 1200 \times (20^2 - 18^2)\)

Equating energy losses to work done against resistance:

\(1200g \times s \times \sin 1^{\circ} + \frac{1}{2} \times 1200 \times (20^2 - 18^2) = 350s\)

Solving gives \(s = 324 \text{ m}\).