(i) The normal reaction force \(R\) is given by \(R = mg \cos \alpha\). Given \(R = 2.4\) N and \(m = 0.25\) kg, we have:

\(2.4 = 0.25g \cos \alpha\)

Solving for \(\cos \alpha\), we get:

\(\cos \alpha = \frac{2.4}{0.25 \times 9.8}\)

\(\cos \alpha = \frac{2.4}{2.45}\)

\(\cos \alpha = 0.9796\)

\(\alpha = \cos^{-1}(0.9796) \approx 16.3^\circ\)

(ii) The coefficient of friction \(\mu\) is given by \(\mu = \tan \alpha\). Therefore:

\(\mu = \tan(16.3^\circ)\)

\(\mu \approx 0.292\)

Thus, the least possible value of \(\mu\) is 0.292.