(i) The function \(f(x) = 5 - 2 \sin 2x\) has a range determined by the range of \(\sin 2x\), which is \([-1, 1]\). Thus, \(f(x)\) ranges from \(5 - 2(1) = 3\) to \(5 - 2(-1) = 7\). Therefore, the range is \(3 \leq f(x) < 7\).

(ii) The graph of \(y = f(x)\) is a sinusoidal curve with one complete oscillation between \(0\) and \(\pi\), initially going downwards, all above \(f(x) = 0\).

(iii) To solve \(f(x) = 6\), set \(5 - 2 \sin 2x = 6\), giving \(\sin 2x = -\frac{1}{2}\). Solving \(2x = \frac{7\pi}{6}\) or \(2x = \frac{11\pi}{6}\), we find \(x = \frac{7\pi}{12}\) or \(x = \frac{11\pi}{12}\).

(iv) For \(g\) to have an inverse, it must be one-to-one. The largest \(k\) for which this is true is \(\frac{\pi}{4}\).

(v) To find \(g^{-1}(x)\), solve \(5 - 2 \sin 2x = y\) for \(x\). This gives \(\sin 2x = \frac{5-y}{2}\), so \(2x = \sin^{-1} \left( \frac{5-y}{2} \right)\) and \(x = \frac{1}{2} \sin^{-1} \left( \frac{5-x}{2} \right)\).