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 \(\mu\):

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

\(\mu = 0.442\)