(i) To find the tension in the string, consider the forces acting on the pulley. The resultant force exerted by the string is \(3\sqrt{3} \text{ N}\). Using the cosine of the angle \(30^\circ\), we have:

\(2T \cos 30^\circ = 3\sqrt{3}\)

\(2T \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\)

\(T \sqrt{3} = 3\sqrt{3}\)

\(T = 3 \text{ N}\)

(ii) For particle Q in limiting equilibrium, resolve forces parallel and perpendicular to AC. The tension \(T = 3 \text{ N}\), and the coefficient of friction \(\mu = 0.75\).

Resolve forces parallel to AC:

\(T = F + mg \sin 30^\circ\)

Resolve forces perpendicular to AC:

\(R = mg \cos 30^\circ\)

Using \(F = \mu R\):

\(3 = 0.75 (10 \cos 30^\circ)m + 10m \sin 30^\circ\)

\(3 = 0.75 \times \frac{\sqrt{3}}{2} \times 10m + 10m \times \frac{1}{2}\)

\(3 = 3.75m \sqrt{3} + 5m\)

Solving for \(m\):

\(m = 0.261 \text{ kg}\)