(i) Using Newton's second law along the plane, the equation is:

\(-F - 0.6g\sin 18^{\circ} = 0.6(-4)\)

Solving for the frictional force \(F\):

\(F = 0.6g\sin 18^{\circ} - 0.6 \times 4\)

\(F = 0.546\, \text{N}\)

For the normal component, resolve forces perpendicular to the plane:

\(R = 0.6g\cos 18^{\circ}\)

\(R = 5.71\, \text{N}\)

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

\(\mu = \frac{F}{R} = \frac{0.546}{5.71} = 0.096\)

(ii) When the particle moves down the plane, using Newton's second law:

\(0.6g\sin 18^{\circ} - 0.546 = 0.6a\)

Solving for \(a\):

\(a = \frac{0.6g\sin 18^{\circ} - 0.546}{0.6}\)

\(a = 2.18\, \text{m s}^{-2}\)