(i) For equilibrium, resolve forces along the plane:

\(100 \cos \theta = 8g \sin 30\)

\(100 \cos \theta = 8 \times 9.8 \times 0.5\)

\(100 \cos \theta = 39.2\)

\(\cos \theta = \frac{39.2}{100}\)

\(\theta = \cos^{-1}(0.392) \approx 66.4^{\circ}\)

Resolve forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 8g \cos 30 + 100 \sin \theta\)

\(R = 8 \times 9.8 \times \frac{\sqrt{3}}{2} + 100 \sin 66.4\)

\(R = 67.84 + 91.8\)

\(R = 161\, \text{N}\)

(ii) Given \(\theta = 30\), find the acceleration \(a\):

Apply Newton's second law parallel to the plane:

\(100 \cos 30 - 8g \sin 30 = 8a\)

\(100 \times \frac{\sqrt{3}}{2} - 8 \times 9.8 \times 0.5 = 8a\)

\(86.6 - 39.2 = 8a\)

\(47.4 = 8a\)

\(a = \frac{47.4}{8} = 5.925\)

Thus, \(a \approx 5.83\, \text{m/s}^2\).