(a) Resolve forces perpendicular to the plane: \(R = 0.3g \cos \theta + 4 \sin \theta = 3 \times \frac{24}{25} + 4 \times \frac{7}{25}\).

Resolve forces parallel to the plane: \(F = 4 \cos \theta - 0.3g \sin \theta = 4 \times \frac{24}{25} - 3 \times \frac{7}{25}\).

Using \(F = \mu R\), solve for \(\mu\): \(3 = \mu \times 4\), hence \(\mu = \frac{3}{4}\).

(b) Use Newton's second law: \(F = \mu \times 0.3g \cos \theta = \frac{3}{4} \times 0.3 \times \frac{24}{25}\).

Net force: \(4 - \frac{3}{4} \times 0.3 \times \frac{24}{25} - 0.3 \times g \times \frac{7}{25} = 0.3a\).

Solve for \(a\): \(a = \frac{10}{3} \text{ m/s}^2\).

(c) Distance in 3 seconds: \(s_1 = \frac{1}{2} \times \frac{10}{3} \times 3^2 = 15 \text{ m}\).

Velocity after 3 seconds: \(v = \frac{10}{3} \times 3 = 10 \text{ m/s}\).

After force is removed, apply Newton's second law: \(-0.3g \sin \theta - \mu \times 0.3g \cos \theta = 0.3a\), leading to \(a = -10 \text{ m/s}^2\).

Use \(v^2 = u^2 + 2as\) to find extra distance \(s_2\): \(0 = 10^2 + 2 \times (-10) \times s_2\), so \(s_2 = 5 \text{ m}\).

Total distance: \(s_1 + s_2 = 15 + 5 = 20 \text{ m}\).