(i) The driving force \(F\) provided by the engine is given by \(F = \frac{1000P}{25}\).

Using Newton's second law, the net force is also given by \(F - 600 = 1000 \times 0.2\).

Substituting the expression for \(F\), we have:

\(\frac{1000P}{25} - 600 = 200\)

\(40P - 600 = 200\)

\(40P = 800\)

\(P = 20\)

(ii) For steady speed, the acceleration \(a = 0\). Therefore, the driving force equals the resistance:

\(\frac{1000P}{v_{\text{max}}} = 600\)

Substituting \(P = 20\), we get:

\(\frac{20000}{v_{\text{max}}} = 600\)

\(v_{\text{max}} = \frac{20000}{600}\)

\(v_{\text{max}} = 33.3 \text{ m s}^{-1}\)