(i) Since the surface is smooth, there is no friction. The block is in equilibrium, so the horizontal components of the forces must balance. The horizontal component of the \(40 \text{ N}\) force is \(40 \cos(90^\circ) = 0\). The horizontal component of the \(X \text{ N}\) force is \(X \cos(30^\circ)\). Therefore, \(X \cos(30^\circ) = 0\), which implies \(X = 20 \text{ N}\).

(ii) For the rough surface, the block is in limiting equilibrium, so the frictional force \(F = 10 \text{ N}\) acts towards \(A\). The normal reaction \(R\) is equal to the weight of the block, \(R = 15g\). The frictional force is given by \(F = \mu R\), so \(10 = \mu \times 15g\). Solving for \(\mu\), we get \(\mu = \frac{10}{15g} = 0.5\).