(i) Resolve forces for particle A:

The normal reaction force is given by:

\(R = mg \cos 30^{\circ}\)

The frictional force is:

\(F = 0.2 \times mg \cos 30^{\circ}\)

For particle B, the tension is:

\(T = 4g\)

For particle A, resolve parallel to the plane:

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

Substitute for T and F:

\(40 = mg \sin 30^{\circ} + 0.2 \times mg \cos 30^{\circ}\)

Solve for m:

\(m = \frac{40}{5 + \sqrt{3}} = 5.94\)

\((ii) For m = 3, resolve forces and use energy principles:\)

Resolve forces for particle A:

\(R = 3g \cos 30^{\circ}\)

\(F = 0.2 \times 3g \cos 30^{\circ}\)

Apply Newton's Second Law:

\(4g - T = 4a\)

\(T - 3g \sin 30^{\circ} - F = 3a\)

Solve for acceleration a:

\(a = \frac{25}{3 \sqrt{3}} - 7\)

Use the equation:

\(v^2 = u^2 + 2as\)

\(Since initial velocity u = 0, solve for s:\)

\(s = \frac{0.5 \times 2.83}{2} = 0.710 \text{ m}\)