(a) Resolve forces parallel to the plane using Newton's second law:

\(T \cos 26^{\circ} - 8g \sin 18^{\circ} - F = 8 \times 0.2\)

Resolve forces perpendicular to the plane:

\(R + T \sin 26^{\circ} = 8g \cos 18^{\circ}\)

Use the friction equation \(F = 0.65R\) to find an equation in \(T\) only.

Solve for \(T\):

\(T = 64 \text{ N}\)

(b) Use the constant acceleration formula to find the distance \(s\) during the fourth second:

\(s = \frac{1}{2} \times 0.2 \times 4^2 - \frac{1}{2} \times 0.2 \times 3^2\)

or

\(s = \frac{1}{2} (0 + 0.2 \times 4) \times 4 - \frac{1}{2} (0 + 0.2 \times 3) \times 3\)

\(Distance = 0.7 m\)

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

For the car moving up the hill, apply Newton's second law:

\(1200 - 350 - 1250 \times 10 \times 0.05 = 1250a\)

\(a = \frac{225}{1250} = 0.18 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 100\):

\(v^2 = 0 + 2 \times 0.18 \times 100\)

\(v = 6 \text{ m/s}\)

For the car moving down the hill, apply Newton's second law:

\(1200 - 350 + 1250 \times 10 \times 0.05 = 1250a\)

\(a = \frac{1475}{1250} = 1.18 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 100\):

\(v^2 = 0 + 2 \times 1.18 \times 100\)

\(v = 15.4 \text{ m/s}\)

First, calculate the normal reaction force, \(R\), on the block. Since the block is on an inclined plane, \(R = 5g \cos 6^{\circ}\).

Next, calculate the frictional force, \(F\), using the formula \(F = \mu R\), where \(\mu = 0.3\). Thus, \(F = 0.3 \times 5g \cos 6^{\circ}\).

Since the block is moving at constant speed, the tension \(T\) in the rope must balance the component of gravitational force down the plane and the frictional force. Therefore, \(T = 5g \sin 6^{\circ} + F\).

Substitute the values to find \(T\):

\(T = 5g \sin 6^{\circ} + 0.3 \times 5g \cos 6^{\circ}\).

Calculating this gives \(T \approx 20.1 \text{ N}\).

(i) To find the friction force, we consider the component of the gravitational force acting down the slope. The gravitational force component is given by:

\(F = mg \sin \theta\)

where \(m = 0.2\) kg, \(g = 9.8\) m/s², and \(\theta = 20^{\circ}\).

Substituting the values, we get:

\(F = 0.2 \times 9.8 \times \sin 20^{\circ} = 0.684\) N

Thus, the friction force is 0.684 N.

(ii) To find the acceleration, we first calculate the normal reaction force \(R\):

\(R = mg \cos \theta = 0.2 \times 9.8 \times \cos 20^{\circ}\)

\(R = 0.2 \times 9.8 \times \cos 20^{\circ} \approx 1.84\) N

The frictional force \(F_f\) is given by:

\(F_f = \mu R = 0.6 \times 1.84 = 1.104\) N

Using Newton's second law along the plane:

\(0.9 + 0.2 \times 9.8 \times \sin 20^{\circ} - F_f = 0.2a\)

\(0.9 + 0.684 - 1.104 = 0.2a\)

\(0.48 = 0.2a\)

\(a = \frac{0.48}{0.2} = 2.4\) m/s²

However, according to the mark scheme, the correct acceleration is 2.28 m/s².