(i) (a) Since \(\theta\) is a reflex angle, it is in the 4th quadrant where sine is negative. Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\(\sin^2 \theta = 1 - \cos^2 \theta = 1 - k^2\)

\(\sin \theta = -\sqrt{1-k^2}\)

(i) (b) \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\sqrt{1-k^2}}{k}\)

(ii) Since \(\theta\) is in the 4th quadrant, \(2\theta\) will be in the range of 540° to 720°, which corresponds to the 3rd and 4th quadrants. In both these quadrants, \(\sin 2\theta\) is negative.