To solve for the greatest possible value of \(P\), we resolve the forces parallel and perpendicular to the plane.

1. Resolve forces perpendicular to the plane:

\(12g \cos 25 = R + P \sin 8\)

2. Resolve forces parallel to the plane:

\(P \cos 8 = F + 12g \sin 25\)

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

\(F = 0.3R\)

4. Substitute \(F = 0.3R\) into the parallel force equation:

\(P \cos 8 = 0.3R + 12g \sin 25\)

5. Substitute \(R = 12g \cos 25 - P \sin 8\) from the perpendicular equation into the parallel equation:

\(P \cos 8 = 0.3(12g \cos 25 - P \sin 8) + 12g \sin 25\)

6. Solve for \(P\):

\(P = 80.8\)

Resolve forces perpendicular to the plane:

\(T \sin 60 + R = 25 \cos 20\)

Resolve forces parallel to the plane:

\(T \cos 60 = F + 25 \sin 20\)

\(T \cos 60 + F = 25 \sin 20\)

Using the friction formula \(F = \mu R\):

\(T \cos 60 = 25 \sin 20 + 0.3(25 \cos 20 - T \sin 60)\)

Substitute \(F = 0.3R\) into the equation:

\(T = \frac{25 \sin 20 + 0.3 \times 25 \cos 20}{\cos 60 + 0.3 \sin 60}\)

Calculate the values:

\(T = 6.26\) N (least value)

\(T = 20.5\) N (greatest value)

(i) The normal reaction force \(R\) is given by:

\(R = 3g \cos 60^{\circ}\)

Using the equation for frictional force \(F = \mu R\), we resolve forces parallel to the plane:

\(3g \sin 60^{\circ} - \mu (3g \cos 60^{\circ}) - 15 = 0\)

Solving for \(\mu\), we find:

\(\mu = 0.732\)

(ii) The maximum force is given by:

\(3g \sin 60^{\circ} + F = 3 \sin 60^{\circ} + \mu (3 \cos 60^{\circ})\)

Solving for \(X\), we find:

\(X = 37.0\)

First, calculate the normal reaction \(R\) on the particle:

\(R = 3g \cos 20^{\circ}\)

Next, calculate the frictional force \(F\) using the coefficient of friction \(\mu = 0.35\):

\(F = 0.35 \times 3g \cos 20^{\circ}\)

For the least possible value of \(P\), resolve forces parallel to the plane:

\(P_1 + F = 3g \sin 20^{\circ}\)

Solving for \(P_1\):

\(P_1 = 3g \sin 20^{\circ} - F\)

Substitute the values to find \(P_1 = 0.394\) (as given).

For the greatest possible value of \(P\), resolve forces in the opposite direction:

\(P_2 = F + 3g \sin 20^{\circ}\)

Substitute the values to find \(P_2 = 20.1 \text{ N}\).

First, resolve the forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 20g \cos 60^{\circ} = 100\).

The frictional force \(F\) is given by \(F = \mu R = 100\mu\).

Resolve the forces parallel to the plane for maximum and minimum values of \(P\):

For maximum \(P\), \(P_{\text{max}} = 20g \sin 60^{\circ} + F\).

For minimum \(P\), \(P_{\text{min}} = 20g \sin 60^{\circ} - F\).

Given \(P_{\text{max}} = 2P_{\text{min}}\), substitute the expressions:

\(20g \sin 60^{\circ} + F = 2(20g \sin 60^{\circ} - F)\).

Simplify and solve for \(\mu\):

\(20g \sin 60^{\circ} + 100\mu = 40g \sin 60^{\circ} - 200\mu\).

\(300\mu = 20g \sin 60^{\circ}\).

\(\mu = \frac{20g \sin 60^{\circ}}{300} = \frac{\sqrt{3}}{3} = 0.577\).