(i) For particle A in limiting equilibrium, the tension \(T\) in the string is equal to the weight of A, so \(T = 0.2g\).

For particle B, the tension \(T\) is equal to the frictional force \(F\), so \(T = F\).

The normal reaction \(R\) on B is \(R = 0.3g\).

Using the friction formula \(F = \mu R\), we have \(0.2g = \mu \times 0.3g\).

Solving for \(\mu\), we get \(\mu = \frac{2}{3}\).

(ii) With the additional force on B, the vertical component is 1.8 N upwards, so the new normal reaction is \(R = 0.3g - 1.8\).

The frictional force is \(F = \frac{2}{3}(0.3g - 1.8)\).

In limiting equilibrium, the horizontal force \(X\) is equal to the sum of the tension and the frictional force, so \(X = T + F\).

Substituting \(T = 0.2g\) and \(F = \frac{2}{3}(0.3g - 1.8)\), we find \(X = 2.8\).