(i) Resolve forces perpendicular to the plane: \(R = 13 \cos 67.4 = 13 \times \frac{5}{13} = 5\). Resolve forces parallel to the plane: \(F + 13 \sin 67.4 = F + 13 \times \frac{12}{13} = 20\). Using \(F = \mu R\), we have \(\mu = \frac{8}{5} = 1.6\).

(ii) Resolve forces parallel to the plane: \(13 \sin 67.4 - F = 1.3a\). Using \(F = \mu R = 8\), we find \(a = 3.08 \text{ m/s}^2\).

(iii) Use \(s = ut + \frac{1}{2}at^2\) with \(u = 0\) and \(a = \frac{40}{13}\) to find the distance moved in the first 2 seconds: \(s = 0 + 0.5 \times \frac{40}{13} \times 2^2 = \frac{80}{13} = 6.15\). Work done \(\text{WD} = F \times d = 8 \times 6.15 = 49.2 \text{ J}\).