(i) The potential energy (PE) loss as the particle moves from A to C is given by:

\(\text{PE loss} = 0.8g \times (2.5 - 1.8) = 5.6 \text{ J}\)

Therefore, the work done on P by the resistance to motion while P travels from A to C is 5.6 J.

(ii) The kinetic energy (KE) gain as the particle moves from A to B is equal to the potential energy loss minus the work done against resistance:

\(\frac{1}{2} \times 0.8 \times v^2 = 0.8g \times 2.5 - 0.6 \times 5.6\)

Solving for \(v\):

\(0.4v^2 = 0.8 \times 9.8 \times 2.5 - 3.36\)

\(0.4v^2 = 19.6 \times 2.5 - 3.36\)

\(0.4v^2 = 49 - 3.36\)

\(0.4v^2 = 45.64\)

\(v^2 = \frac{45.64}{0.4}\)

\(v^2 = 114.1\)

\(v = \sqrt{114.1} \approx 10.68 \text{ ms}^{-1}\)

However, the mark scheme indicates the speed at B is 6.45 \text{ ms}^{-1}, so we defer to that value.