(i) To find the acceleration, we start by calculating the normal reaction force, which is given by:

\(R = mg \cos 25^{\circ}\)

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

\(F = 0.4mg \cos 25^{\circ}\)

Using Newton's Second Law along the plane:

\(mg \sin 25^{\circ} - 0.4mg \cos 25^{\circ} = ma\)

Solving for \(a\):

\(a = g(\sin 25^{\circ} - 0.4 \cos 25^{\circ})\)

Substituting \(g = 9.8 \text{ m/s}^2\), we find:

\(a = 0.601 \text{ m/s}^2\)

(ii) To find the distance travelled in the first 3 seconds, we use the equation:

\(s = ut + \frac{1}{2}at^2\)

Since the initial velocity \(u = 0\), the equation simplifies to:

\(s = \frac{1}{2} \times 0.601 \times 3^2\)

Calculating this gives:

\(s = 2.70 \text{ m}\)