(i) The forces acting on P are:

• Weight \(W = 0.13g\) acting vertically downwards.
• Tension \(T\) in the string acting up the plane.
• Frictional force \(F\) acting down the plane.
• Normal reaction \(R\) perpendicular to the plane.

(ii) To show \(T - F > 0.32\):

The component of weight of P down the plane is \(0.13g \sin \theta\), where \(\sin \theta = \frac{16}{65}\).

Thus, \(0.13g \sin \theta = 0.13g \left(\frac{16}{65}\right)\).

The resultant upward force (RUF) is \(T - F - 0.13g \left(\frac{16}{65}\right)\).

For P to move upwards, \(T - F - 0.13g \left(\frac{16}{65}\right) > 0\).

Therefore, \(T - F > 0.13g \left(\frac{16}{65}\right) \approx 0.32\).

(iii) To find the acceleration of P:

The normal reaction \(R = 0.13g \cos \theta\), where \(\cos \theta = \frac{63}{65}\).

Thus, \(R = 0.13g \left(\frac{63}{65}\right) \approx 1.26\).

The frictional force \(F = \mu R = 0.6 \times 1.26 = 0.756\).

Using Newton's second law for P:

\(T - 0.32 = 0.13a\) and \(0.11g - T = 0.11a\).

Solving these equations gives \(a = 0.1 \text{ m/s}^2\).