(i) In limiting equilibrium, the frictional force is equal to the tension in the string. Therefore, we have:

\(T = \mu R\)

Given that the tension \(T = 24 \text{ N}\) and the weight \(W = 30 \text{ N}\), the normal reaction \(R = W = 30 \text{ N}\). Thus:

\(24 = \mu \times 30\)

Solving for \(\mu\):

\(\mu = \frac{24}{30} = 0.8\)

(ii) When the block is in motion, resolve the forces vertically and horizontally. The vertical component of the tension is \(25 \sin 30^\circ\) and the horizontal component is \(25 \cos 30^\circ\).

The normal reaction \(R\) is given by:

\(R = 30 - 25 \sin 30^\circ\)

The frictional force \(F\) is:

\(F = \mu R = 0.8 \times (30 - 25 \sin 30^\circ) = 14 \text{ N}\)

Using Newton's second law horizontally:

\(25 \cos 30^\circ - F = \frac{30}{g} a\)

Substitute \(F = 14 \text{ N}\) and solve for \(a\):

\(25 \cos 30^\circ - 14 = \frac{30}{g} a\)

\(a = 2.55 \text{ m/s}^2\)