(a)(i) Resolve forces parallel to the plane:

\(T \cos 20^{\circ} - 5 - 2g \sin 30^{\circ} = 2 \times 1.2\)

Solving for T:

\(T \cos 20^{\circ} = 5 + 2g \sin 30^{\circ} + 2 \times 1.2\)

\(T = \frac{5 + 2g \sin 30^{\circ} + 2 \times 1.2}{\cos 20^{\circ}}\)

\(T = 18.5 \text{ N}\)

(a)(ii) Resolve forces perpendicular to the plane:

\(R = 2g \cos 30^{\circ} - T \sin 20^{\circ}\)

\(Using F = μR:\)

\(5 = \mu (2g \cos 30^{\circ} - T \sin 20^{\circ})\)

Solving for μ:

\(\mu = \frac{5}{2g \cos 30^{\circ} - T \sin 20^{\circ}}\)

\(\mu = 0.455\)

(b) Calculate the maximum friction force:

\(\text{Max } F = 0.8 \times (2g \cos 30^{\circ} - 15 \sin 20^{\circ})\)

\(= 9.7521 \ldots\)

Net force up the plane:

\(15 \cos 20^{\circ} - 2g \sin 30^{\circ} = 4.0953 \ldots\)

Since 4.0953 < 9.7521, the block does not move up the plane.