(i) Resolve forces parallel to the slope for Fig. 1. The force equation is:

\(F + W \sin \alpha = 7.2\)

Where \(F \leq \mu R\) and \(R = W \cos \alpha\). Substituting, we have:

\(\mu \times 7.5 \cos \alpha \geq 7.2 - 7.5 \sin \alpha\)

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

Substitute these values:

\(\mu \times 7.5 \times \frac{24}{25} \geq 7.2 - 7.5 \times \frac{7}{25}\)

\(\mu \geq \frac{17}{24}\)

(ii) For Fig. 2, the resultant force down the plane is:

\(7.2 + 7.5 \times \frac{7}{25} - \mu (7.5 \times \frac{24}{25}) > 0\)

Simplifying gives:

\(\mu < \frac{31}{24}\)