To solve the problem, we need to resolve the forces acting on the ring.

(i) The frictional component of the contact force is the horizontal component of the tension in the string. Given \(\tan \alpha = \frac{5}{12}\), we can find \(\cos \alpha\) using the identity \(\cos \alpha = \frac{12}{13}\). Thus, the frictional force \(F = 13 \cos \alpha = 13 \times \frac{12}{13} = 12 \text{ N}\).

(ii) The normal component of the contact force is the sum of the vertical component of the tension and the weight of the ring. The vertical component of the tension is \(13 \sin \alpha\), where \(\sin \alpha = \frac{5}{13}\). Therefore, the normal force \(R = 1.1 \times 10 + 13 \sin \alpha = 11 + 13 \times \frac{5}{13} = 16 \text{ N}\).

(iii) In limiting equilibrium, the frictional force is equal to the product of the coefficient of friction \(\mu\) and the normal force \(R\). Thus, \(\mu = \frac{F}{R} = \frac{12}{16} = 0.75\).