(a) The power \(P\) is given by the formula \(P = Fv\), where \(F\) is the force and \(v\) is the velocity.

Here, \(F = 350 \text{ N}\) and \(v = 20 \text{ m s}^{-1}\).

Thus, \(P = 350 \times 20 = 7000 \text{ W} = 7 \text{ kW}\).

(b) The power \(P\) is 15 kW, which is \(15000 \text{ W}\).

Using \(P = Fv\), we have \(15000 = DF \times 20\), where \(DF\) is the driving force.

Solving for \(DF\), we get \(DF = 750 \text{ N}\).

Using Newton's second law, \(DF - 350 = 1400a\).

Substituting \(DF = 750\), we have \(750 - 350 = 1400a\).

Thus, \(400 = 1400a\).

Solving for \(a\), we get \(a = \frac{400}{1400} = \frac{2}{7} \approx 0.286 \text{ m s}^{-2}\).