(i) Resolve the forces horizontally: \(5 \cos \alpha = F\). Given \(\tan \alpha = \frac{3}{4}\), \(\cos \alpha = \frac{4}{5}\). Thus, \(F = 5 \times \frac{4}{5} = 4 \text{ N}\).

Resolve the forces vertically: \(R + 5 \sin \alpha = 8\). Since \(\sin \alpha = \frac{3}{5}\), \(R + 5 \times \frac{3}{5} = 8\), so \(R = 8 - 3 = 5 \text{ N}\).

Using \(F = \mu R\), we have \(4 = 5 \mu\). Therefore, \(\mu = 0.8\).

(ii) Resolve the forces vertically with the new force: \(R + 10 \sin \alpha = 8\). \(R + 10 \times \frac{3}{5} = 8\), so \(R = 8 - 6 = 2 \text{ N}\).

Using \(F = \mu R\), \(F = 0.8 \times 2 = 1.6 \text{ N}\).

Resolve horizontally: \(10 \cos \alpha - F = 0.8a\). \(10 \times \frac{4}{5} - 1.6 = 0.8a\), so \(8 - 1.6 = 0.8a\).

Thus, \(6.4 = 0.8a\) and \(a = 8 \text{ ms}^{-2}\).