(i) Using the work-energy principle, the change in kinetic energy is equal to the work done against resistance:

\(\frac{1}{2} \times 1.2 \times 7.5^2 - \frac{1}{2} \times 1.2 \times v^2 = 25\)

Solving for \(v\):

\(\frac{1}{2} \times 1.2 \times 7.5^2 = 33.75\)

\(33.75 - 25 = \frac{1}{2} \times 1.2 \times v^2\)

\(8.75 = 0.6v^2\)

\(v^2 = \frac{8.75}{0.6}\)

\(v = \sqrt{14.5833} \approx 3.82 \text{ m s}^{-1}\)

(ii) Considering the inclined plane, the potential energy change is \(1.2g \sin 30 \times d\).

Using the work-energy principle:

\(\frac{1}{2} \times 1.2 \times 7.5^2 - 25 + 1.2g \sin 30 \times d = \frac{1}{2} \times 1.2 \times 9^2\)

\(33.75 - 25 + 1.2 \times 9.81 \times 0.5 \times d = 48.6\)

\(8.75 + 5.886d = 48.6\)

\(5.886d = 39.85\)

\(d = \frac{39.85}{5.886} \approx 6.64 \text{ m}\)