(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.