(a) Resolving forces vertically and horizontally:

Normal reaction, \(R = T \sin 30^\circ + 0.1g\).

Frictional force, \(F = T \cos 30^\circ\).

For equilibrium, \(T \cos 30^\circ = 0.8 (T \sin 30^\circ + 0.1g)\).

Solving for \(T\):

\(T \cos 30^\circ = 0.8T \sin 30^\circ + 0.08g\).

\(T (\cos 30^\circ - 0.8 \sin 30^\circ) = 0.08g\).

\(T = \frac{0.08g}{\cos 30^\circ - 0.8 \sin 30^\circ}\).

\(T = 1.72 \text{ N}\).

(b) When \(T = 3\):

Normal reaction, \(R = 3 \sin 30^\circ + 0.1g\).

Net force, \(3 \cos 30^\circ - 0.8(3 \sin 30^\circ + 0.1g) = 0.1a\).

Solving for \(a\):

\(a = \frac{3 \cos 30^\circ - 0.8(3 \sin 30^\circ + 0.1g)}{0.1}\).

\(a = 5.98 \text{ m/s}^2\).