(i) The power exerted by the runner is given by \(P = Fv\), where \(F\) is the force. On horizontal ground, the resistance force \(F = kv\). Therefore, \(P = kv^2\).

Given \(P = 100 \text{ W}\) and \(v = 4 \text{ m s}^{-1}\), we have:

\(100 = k \times 4^2\)

\(100 = 16k\)

\(k = \frac{100}{16} = 6.25\)

(ii) When running uphill, the forces acting on the runner include the resistance \(kv\), gravitational component \(70g \sin \alpha\), and the power equation \(P = Fv\).

The equation becomes:

\(\frac{100}{v} = 70g \times 0.05 + 6.25v\)

\(\frac{100}{v} = 35 + 6.25v\)

Multiplying through by \(v\) gives:

\(100 = 35v + 6.25v^2\)

Rearranging gives the quadratic equation:

\(6.25v^2 + 35v - 100 = 0\)

Solving this quadratic equation using the quadratic formula \(v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6.25\), \(b = 35\), and \(c = -100\), we find:

\(v = \frac{-35 \pm \sqrt{35^2 - 4 \times 6.25 \times (-100)}}{2 \times 6.25}\)

\(v = \frac{-35 \pm \sqrt{1225 + 2500}}{12.5}\)

\(v = \frac{-35 \pm \sqrt{3725}}{12.5}\)

Calculating gives the maximum speed \(v \approx 2.08 \text{ m s}^{-1}\).