(i) The potential energy (PE) at A is given by \(0.8g \times 4\), where \(g\) is the acceleration due to gravity. This energy is converted into kinetic energy (KE) at C. Therefore, \(\frac{1}{2} \times 0.8 \times v^2 = 0.8g \times 4\). Solving for \(v\), we get \(v = 8.94 \text{ ms}^{-1}\).

(ii) The force of friction \(F\) is \(0.3 \times 0.8g\). The work done against friction over the distance BC is \(F \times 5\). The loss in potential energy is \(0.8g \times 4\). The kinetic energy at C is given by \(\frac{1}{2} \times 0.8 \times v^2 = 0.8g \times 4 - F \times 5\). Solving for \(v\), we get \(v = 7.07 \text{ ms}^{-1}\).