To find the coefficient of friction, we resolve the forces vertically and horizontally.

Vertically, the normal reaction force is given by:

\(R + 2000 \cos 15^{\circ} = 400g\)

where \(g\) is the acceleration due to gravity \((9.8 \text{ m/s}^2)\).

Horizontally, the frictional force \(F\) is:

\(F = 2000 \sin 15^{\circ}\)

In limiting equilibrium, the frictional force \(F\) is equal to \(\mu R\), where \(\mu\) is the coefficient of friction:

\(2000 \sin 15^{\circ} = \mu (400g - 2000 \cos 15^{\circ})\)

Solving for \(\mu\):

\(\mu = \frac{2000 \sin 15^{\circ}}{400g - 2000 \cos 15^{\circ}}\)

Substitute \(g = 9.8 \text{ m/s}^2\):

\(\mu = \frac{2000 \times 0.2588}{400 \times 9.8 - 2000 \times 0.9659}\)

\(\mu = \frac{517.6}{3920 - 1931.8}\)

\(\mu = \frac{517.6}{1988.2}\)

\(\mu \approx 0.26\)

Rounding to 2 significant figures, the coefficient of friction is \(0.25\).

(i) Resolve horizontally:

\(T \cos 60^\circ = 75 \cos 30^\circ\)

\(T \times \frac{1}{2} = 75 \times \frac{\sqrt{3}}{2}\)

\(T = 75 \sqrt{3} \approx 130 \text{ N}\)

Resolve vertically:

\(T \sin 60^\circ + 75 \sin 30^\circ + R = 20g\)

\(130 \times \frac{\sqrt{3}}{2} + 75 \times \frac{1}{2} + R = 200\)

\(R = 200 - 130 \times \frac{\sqrt{3}}{2} - 75 \times \frac{1}{2}\)

\(R \approx 50 \text{ N}\)

(ii) Resolve horizontally:

\(T \cos 60^\circ + 25 = 75 \cos 30^\circ\)

\(T \times \frac{1}{2} + 25 = 75 \times \frac{\sqrt{3}}{2}\)

\(T = 79.9 \text{ N}\)

Resolve vertically:

\(79.9 \sin 60^\circ + 75 \sin 30^\circ + R = 200\)

\(R = 200 - 79.9 \times \frac{\sqrt{3}}{2} - 75 \times \frac{1}{2}\)

\(R \approx 93.3 \text{ N}\)

Coefficient of friction:

\(\mu = \frac{25}{93.3} \approx 0.268\)

(a) In limiting equilibrium, the frictional force is equal to the applied force. The normal reaction \(R\) is given by \(R = 0.2g\).

The frictional force is \(\mu R\), so \(1.2 = \mu \times 0.2g\).

Solving for \(\mu\), we get \(\mu = \frac{1.2}{0.2g} = 0.6\).

(b) Using Newton's Second Law, the net force is \(1.2 - 0.3 \times 0.2g = 0.2a\).

Solving for \(a\), we get \(a = 3 \, \text{m/s}^2\).

To find the distance travelled in the third second, use the equation \(s = ut + \frac{1}{2}at^2\).

For the first 3 seconds, \(s_3 = 0 + \frac{1}{2} \times 3 \times 3^2 = 13.5 \, \text{m}\).

For the first 2 seconds, \(s_2 = 0 + \frac{1}{2} \times 3 \times 2^2 = 6 \, \text{m}\).

Thus, the distance travelled in the third second is \(13.5 - 6 = 7.5 \, \text{m}\).

(i) Using the equation of motion, the acceleration is given by:

\(v = u + at\)

\(3.5 = 0 + 10a\)

\(a = 0.35 \text{ m/s}^2\)

Applying Newton's second law horizontally:

\(10\cos(15^{\circ}) - F = 2 \times 0.35\)

\(F = 10\cos(15^{\circ}) - 0.7\)

\(F = 8.96 \text{ N}\)

(ii) Resolving forces vertically, the normal reaction \(R\) is:

\(R = 2g - 10\sin(15^{\circ})\)

The coefficient of friction \(\mu\) is given by:

\(F = \mu R\)

\(\mu = \frac{8.96}{2g - 10\sin(15^{\circ})}\)

\(\mu = 0.515\)

(i) Resolve the forces horizontally: \(5 \cos \alpha = F\). Given \(\tan \alpha = \frac{3}{4}\), \(\cos \alpha = \frac{4}{5}\). Thus, \(F = 5 \times \frac{4}{5} = 4 \text{ N}\).

Resolve the forces vertically: \(R + 5 \sin \alpha = 8\). Since \(\sin \alpha = \frac{3}{5}\), \(R + 5 \times \frac{3}{5} = 8\), so \(R = 8 - 3 = 5 \text{ N}\).

Using \(F = \mu R\), we have \(4 = 5 \mu\). Therefore, \(\mu = 0.8\).

(ii) Resolve the forces vertically with the new force: \(R + 10 \sin \alpha = 8\). \(R + 10 \times \frac{3}{5} = 8\), so \(R = 8 - 6 = 2 \text{ N}\).

Using \(F = \mu R\), \(F = 0.8 \times 2 = 1.6 \text{ N}\).

Resolve horizontally: \(10 \cos \alpha - F = 0.8a\). \(10 \times \frac{4}{5} - 1.6 = 0.8a\), so \(8 - 1.6 = 0.8a\).

Thus, \(6.4 = 0.8a\) and \(a = 8 \text{ ms}^{-2}\).