(i) Resolve the forces vertically. The normal force \(N\) and the vertical component of the force \(X \cos \theta\) balance the weight of the slab:

\(N + X \cos \theta = mg\)

Given \(\tan \theta = \frac{7}{24}\), we find \(\cos \theta = \frac{24}{25}\).

Substitute \(m = 320 \text{ kg}\) and \(g = 10 \text{ m/s}^2\):

\(N + X \left( \frac{24}{25} \right) = 320 \times 10\)

\(N = 3200 - \frac{24}{25}X\)

(ii) Resolve the forces horizontally. The frictional force \(F\) is given by \(F = X \sin \theta\).

Given \(\sin \theta = \frac{7}{25}\), the frictional force is:

\(F = X \left( \frac{7}{25} \right)\)

The slab is about to slip when \(F = \mu N\), where \(\mu = \frac{3}{8}\):

\(\frac{7}{25}X = \frac{3}{8} \left( 3200 - \frac{24}{25}X \right)\)

Solving for \(X\):

\(\frac{7}{25}X = \frac{3}{8} \times 3200 - \frac{3}{8} \times \frac{24}{25}X\)

\(\frac{7}{25}X + \frac{72}{200}X = 1200\)

\(\frac{175}{625}X + \frac{72}{200}X = 1200\)

\(\frac{187}{625}X = 1200\)

\(X = 1875\)