(i) We have \(a = \\sin \theta - 3 \\cos \theta\) and \(b = 3 \\sin \theta + \\cos \theta\).

Calculate \(a^2 + b^2\):

\(a^2 = (\\sin \theta - 3 \\cos \theta)^2 = \\sin^2 \theta - 6 \\sin \theta \\cos \theta + 9 \\cos^2 \theta\)

\(b^2 = (3 \\sin \theta + \\cos \theta)^2 = 9 \\sin^2 \theta + 6 \\sin \theta \\cos \theta + \\cos^2 \theta\)

Adding these, \(a^2 + b^2 = \\sin^2 \theta - 6 \\sin \theta \\cos \theta + 9 \\cos^2 \theta + 9 \\sin^2 \theta + 6 \\sin \theta \\cos \theta + \\cos^2 \theta\)

\(= 10 \\sin^2 \theta + 10 \\cos^2 \theta\)

Using \(\\sin^2 \theta + \\cos^2 \theta = 1\), we get \(a^2 + b^2 = 10\).

(ii) Given \(2a = b\), substitute \(a\) and \(b\):

\(2(\\sin \theta - 3 \\cos \theta) = 3 \\sin \theta + \\cos \theta\)

\(2 \\sin \theta - 6 \\cos \theta = 3 \\sin \theta + \\cos \theta\)

\(-\\sin \theta = 7 \\cos \theta\)

\(\\tan \theta = -\frac{1}{7}\)

\(\theta = \\arctan(-\frac{1}{7}) \approx 98.1^\circ\) and \(\theta = 98.1^\circ + 180^\circ = 278.1^\circ\).