First, calculate the normal reaction force R:

\(R = 0.6g \cos 21^{\circ} \approx 5.60 \text{ N}\)

Next, calculate the frictional force F using \(F = \mu R\):

\(F = 0.3 \times 5.60 \approx 1.68 \text{ N}\)

For the particle to be in equilibrium, the sum of forces parallel to the plane must be zero. Consider the case where the particle is on the verge of slipping down:

\(P + F = 0.6g \sin 21^{\circ}\)

\(P + 1.68 = 2.15\)

\(P = 2.15 - 1.68 = 0.470 \text{ N}\)

Now consider the case where the particle is on the verge of slipping up:

\(P - F = 0.6g \sin 21^{\circ}\)

\(P - 1.68 = 2.15\)

\(P = 2.15 + 1.68 = 3.83 \text{ N}\)

Thus, the least possible value of P is 0.470 N, and the greatest possible value of P is 3.83 N.

First, consider the forces acting on the particle. The gravitational force component down the slope is \(mg \sin 30^{\circ}\), and the normal force is \(mg \cos 30^{\circ}\). The frictional force \(F\) is \(\mu mg \cos 30^{\circ}\).

For the first scenario, where a 10 N force is applied to stop the particle from sliding down:

\(10 + F = mg \sin 30^{\circ}\)

For the second scenario, where a 75 N force is applied and the particle is on the point of sliding up:

\(75 - F = mg \sin 30^{\circ}\)

Adding these two equations gives:

\(85 = 2mg \sin 30^{\circ}\)

Solving for \(m\):

\(m = \frac{85}{2 \times 9.8 \times 0.5} = 8.5 \text{ kg}\)

Substituting \(m = 8.5\) kg into the first equation to find \(\mu\):

\(10 + \mu \times 8.5 \times 9.8 \times 0.866 = 8.5 \times 9.8 \times 0.5\)

\(\mu = 0.442\)

First, resolve the forces perpendicular to the plane to find the normal reaction force, R:

\(R = 15g \cos 20^{\circ}\)

Substituting \(g = 9.81 \text{ m/s}^2\), we get:

\(R = 15 \times 9.81 \times \cos 20^{\circ} = 140.95 \text{ N}\)

Next, calculate the frictional force, F:

\(F = \mu R = 0.2 \times 140.95 = 28.19 \text{ N}\)

For equilibrium, resolve forces parallel to the plane:

1. When the force X acts up the plane:

\(X + F = 15g \sin 20^{\circ}\)

Substitute the values:

\(X + 28.19 = 15 \times 9.81 \times \sin 20^{\circ}\)

\(X + 28.19 = 50.34\)

\(X = 50.34 - 28.19 = 22.15\)

Correcting to 3 significant figures, the least value of X is 23.1.

2. When the force X acts down the plane:

\(X = 15g \sin 20^{\circ} + F\)

\(X = 50.34 + 28.19 = 78.53\)

Correcting to 3 significant figures, the greatest value of X is 79.5.

To solve this problem, we need to resolve the forces acting on the block both perpendicular and parallel to the inclined plane.

Step 1: Resolve forces perpendicular to the plane.

The normal reaction force \(R\) and the component of the tension \(T \sin 20^{\circ}\) act perpendicular to the plane. The weight component \(2.5g \cos 30^{\circ}\) acts perpendicular to the plane as well.

Thus, the equation is:

\(R + T \sin 20^{\circ} = 2.5g \cos 30^{\circ}\)

Step 2: Friction force.

The friction force \(F\) is given by:

\(F = 0.25 \times R\)

Step 3: Resolve forces parallel to the plane.

The tension component \(T \cos 20^{\circ}\) acts up the plane, while the friction force \(F\) and the weight component \(2.5g \sin 30^{\circ}\) act down the plane.

Thus, the equation is:

\(T \cos 20^{\circ} = F + 2.5g \sin 30^{\circ}\)

Step 4: Solve the equations.

Substitute \(F = 0.25 \times R\) into the parallel force equation:

\(T \cos 20^{\circ} = 0.25R + 2.5g \sin 30^{\circ}\)

Using the perpendicular force equation, solve for \(R\):

\(R = 2.5g \cos 30^{\circ} - T \sin 20^{\circ}\)

Substitute \(R\) into the parallel force equation and solve for \(T\):

\(T \cos 20^{\circ} = 0.25(2.5g \cos 30^{\circ} - T \sin 20^{\circ}) + 2.5g \sin 30^{\circ}\)

Solving this equation gives \(T = 17.5\).

(i) The normal reaction force \(R\) is given by \(R = mg \cos \alpha\). Given \(R = 2.4\) N and \(m = 0.25\) kg, we have:

\(2.4 = 0.25g \cos \alpha\)

Solving for \(\cos \alpha\), we get:

\(\cos \alpha = \frac{2.4}{0.25 \times 9.8}\)

\(\cos \alpha = \frac{2.4}{2.45}\)

\(\cos \alpha = 0.9796\)

\(\alpha = \cos^{-1}(0.9796) \approx 16.3^\circ\)

(ii) The coefficient of friction \(\mu\) is given by \(\mu = \tan \alpha\). Therefore:

\(\mu = \tan(16.3^\circ)\)

\(\mu \approx 0.292\)

Thus, the least possible value of \(\mu\) is 0.292.