(a) Resolve forces for the 5 kg particle: The normal reaction is given by the weight of the 5 kg particle, so \(R = 5g\). The frictional force \(F\) is the difference between the weight of the 6 kg particle and the 4 kg particle, so \(F = 6g - 4g = 2g\). The coefficient of friction \(\mu\) is given by \(\mu = \frac{F}{R} = \frac{2g}{5g} = 0.4\).

(b) Apply Newton's second law to the 4 kg and 8 kg particles: For the 4 kg particle, \(T_1 - 4g = 4a\). For the 8 kg particle, \(8g - T_2 = 8a\). The tension difference is \(T_2 - T_1 = F = 0.4 \times 5g\). Adding the equations gives \(8g - 4g - 2g = 17a\), leading to \(a = 1.18 \text{ m/s}^2\). Solving for tensions, \(T_1 = 44.7 \text{ N}\) and \(T_2 = 70.6 \text{ N}\).

(c) For the 8 kg particle, \(T - 4g = 4a\) and \(-T - F = 5a\), or \(-4g - 2g = 9a\). Solving gives \(a = \frac{60}{9}\). Use \(v^2 = u^2 + 2as\) to find \(v\) when the 8 kg particle reaches the ground: \(v^2 = 2 \times \frac{20}{17} \times 0.5 = \frac{20}{17}\), leading to \(v = 1.0846 \ldots\). Use \(v = u + at\) to find \(t\): \(0 = \frac{20}{17} - \frac{60}{9}t\), giving \(t = 0.163 \text{ s}\).