(i) The acceleration for \(t < 0.8\) is calculated using the formula \(a = \frac{v}{t} = \frac{4}{0.8} = 5 \text{ m s}^{-2}\). Since the plane is smooth between A and B, the acceleration is due to gravity: \(a = g \sin \theta\). Therefore, \(5 = 10 \sin \theta\), giving \(\sin \theta = 0.5\). Thus, \(\theta = 30^\circ\).

(ii) For \(0.8 < t < 4.8\), the acceleration is \(\frac{-4}{4.8 - 0.8} = -1 \text{ m s}^{-2}\). Using Newton's second law, \(mg \sin 30^\circ - F = m(-1)\), where \(F\) is the frictional force. The frictional force \(F = \mu R\), where \(R = mg \cos 30^\circ\). Therefore, \(\mu = \frac{mg \sin 30^\circ + m}{mg \cos 30^\circ}\). Simplifying gives \(\mu = 0.693\).