(i) Resolve forces parallel to the plane: \(W \sin \alpha + F = 40\).

Calculate the component of weight parallel to the plane: \(F = 40 - 300 \times 0.1 = 10 \text{ N}\).

Calculate the normal reaction: \(R = 300 \sqrt{1 - 0.1^2} = 298.496 \ldots \text{ N}\).

Use the formula for the coefficient of friction: \(\mu = \frac{F}{R} = \frac{10}{298.496} \approx 0.0335\).

(ii) The component of weight (30 N) is greater than the frictional force (10 N), so the box does not remain in equilibrium.

Given \(\tan \alpha = 2.4\), we find \(\sin \alpha = \frac{12}{13}\) and \(\cos \alpha = \frac{5}{13}\) using the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\).

Resolving forces parallel to the plane:

\(0.6g \sin \alpha = F + P \cos \alpha\)

Resolving forces perpendicular to the plane:

\(R = 0.6g \cos \alpha + P \sin \alpha\)

Using \(F = \mu R\):

\(0.6g \sin \alpha - P \cos \alpha = 0.4(0.6g \cos \alpha + P \sin \alpha)\)

Substituting \(g = 9.8\), \(\sin \alpha = \frac{12}{13}\), and \(\cos \alpha = \frac{5}{13}\):

\(6 \times \frac{12}{13} - P \times \frac{5}{13} = 2.4 \times \frac{5}{13} + 0.4P \times \frac{12}{13}\)

Simplifying gives:

\(6(12/13) - P(5/13) = 2.4(5/13) + 0.4P(12/13)\)

Solving for \(P\):

\(P = 6.12\)

To find the set of possible values of \(P\), we need to consider the forces acting on the particle and the conditions for equilibrium.

1. Resolve forces parallel to the plane:

\(P = \pm F + 0.6g \sin 25^{\circ}\)

where \(F\) is the frictional force.

2. The maximum frictional force \(F\) is given by:

\(F = \mu R\)

where \(R\) is the normal reaction force.

3. Resolve forces perpendicular to the plane to find \(R\):

\(R = 0.6g \cos 25^{\circ}\)

4. Substitute \(R\) into the friction formula:

\(F = 0.36 \times 0.6g \cos 25^{\circ}\)

5. Calculate \(P_{\text{max}}\) and \(P_{\text{min}}\):

\(P_{\text{max}} = F + 0.6g \sin 25^{\circ}\)

\(P_{\text{min}} = -F + 0.6g \sin 25^{\circ}\)

6. Substitute the values to find \(P_{\text{max}}\) and \(P_{\text{min}}\):

\(P_{\text{max}} = 0.36 \times 0.6g \cos 25^{\circ} + 0.6g \sin 25^{\circ}\)

\(P_{\text{min}} = -0.36 \times 0.6g \cos 25^{\circ} + 0.6g \sin 25^{\circ}\)

7. Calculate the numerical values:

\(P_{\text{max}} = 4.49\)

\(P_{\text{min}} = 0.578\)

Therefore, the set of possible values for \(P\) is \(\{ P : 0.578 \leq P \leq 4.49 \}\).

(i) Resolve forces perpendicular to the plane:

\(R + 0.6 \sin \alpha = 0.5g \cos \alpha\)

Given \(\sin \alpha = 0.28\), \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - 0.28^2} = 0.96\).

\(R + 0.6 \times 0.28 = 0.5 \times 9.8 \times 0.96\)

\(R + 0.168 = 4.704\)

\(R = 4.704 - 0.168 = 4.536\)

Normal component is approximately 4.63 N.

(ii) Resolve forces parallel to the line of greatest slope:

\(F + 0.6 \cos \alpha = 0.5g \sin \alpha\)

\(F + 0.6 \times 0.96 = 0.5 \times 9.8 \times 0.28\)

\(F + 0.576 = 1.372\)

\(F = 1.372 - 0.576 = 0.796\)

Frictional component is approximately 0.824 N.

(iii) Coefficient of friction \(\mu\):

\(\mu = \frac{F}{R} = \frac{0.824}{4.63}\)

\(\mu \approx 0.178\)

Resolve forces parallel to the plane:

\(65 \cos 36^{\circ} = 12g \sin 24^{\circ} + F\)

\(F = 65 \cos 36^{\circ} - 12g \sin 24^{\circ} = 3.777707 \ldots\)

Resolve forces perpendicular to the plane:

\(12g \cos 24^{\circ} = R + 65 \sin 36^{\circ}\)

\(R = 12g \cos 24^{\circ} - 65 \sin 36^{\circ} = 71.419 \ldots\)

Use \(F = \mu R\):

\(\mu = \frac{F}{R} = \frac{3.777707}{71.419} = 0.0529\)