(i) Resolve forces along the plane for particle A:

\(T - 0.8g \sin 30^{\circ} = 0.8a\)

Resolve forces along the plane for particle B:

\(1.2g \sin 60^{\circ} - T = 1.2a\)

Combine the equations:

\(1.2g \sin 60^{\circ} - 0.8g \sin 30^{\circ} = 2a\)

Solving for \(a\):

\(a = 3\sqrt{3} - 2 = 3.20 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = \frac{12}{5}(1 + \sqrt{3}) = 6.56 \text{ N}\)

(ii) Normal reactions:

\(R_A = 0.8g \cos 30^{\circ} = 4\sqrt{3}\)

\(R_B = 1.2g \cos 60^{\circ} = 6\)

Frictional forces:

\(F_A = 4\sqrt{3} \mu\)

\(F_B = 6\mu\)

Resolve parallel to the plane:

\(12 \sin 60^{\circ} - 6\mu - T = 0\)

or

\(T - 8 \sin 30^{\circ} - 4\sqrt{3} \mu = 0\)

System equation:

\(12 \sin 60^{\circ} - 8 \sin 30^{\circ} - 6\mu - 4\sqrt{3} \mu = 0\)

Solve for \(\mu\):

\(\mu = \frac{6\sqrt{3} - 4}{6 + 4\sqrt{3}} = 0.494\)