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

(ii) The graph of \(y = 3 - 2 \sin x\) is a sinusoidal wave starting at \(y = 3\), reaching a minimum of \(y = 1\) and a maximum of \(y = 5\), completing one full oscillation from \(x = 0^\circ\) to \(x = 360^\circ\).

(iii) For \(g\) to have an inverse, it must be one-to-one. The largest interval for \(x\) where \(\sin x\) is one-to-one is \(0^\circ \leq x \leq 90^\circ\) or \(0 \leq x \leq \frac{\pi}{2}\).

(iv) To find \(g^{-1}(x)\), set \(y = 3 - 2 \sin x\) and solve for \(x\):

\(y = 3 - 2 \sin x\)

\(2 \sin x = 3 - y\)

\(\sin x = \frac{3 - y}{2}\)

\(x = \sin^{-1}\left(\frac{3 - y}{2}\right)\)

Thus, \(g^{-1}(x) = \sin^{-1}\left(\frac{3-x}{2}\right)\).

(a) The equation of the curve is \(y = 2 + 3 \, \sin \frac{1}{2}x\). The sine function \(\sin \frac{1}{2}x\) oscillates between -1 and 1. Therefore, the maximum value of \(y\) is when \(\sin \frac{1}{2}x = 1\), giving \(y = 2 + 3 \times 1 = 5\). The minimum value of \(y\) is when \(\sin \frac{1}{2}x = -1\), giving \(y = 2 + 3 \times (-1) = -1\).

(b) The sketch of the curve should show one complete cycle starting and finishing at \(y = 2\), with maximum at \(y = 5\) and minimum at \(y = -1\). The curve should have the correct shape and curvature.

(c) To find the number of solutions for \(2 + 3 \, \sin \frac{1}{2}x = 5 - 2x\), rearrange to \(3 \, \sin \frac{1}{2}x = 3 - 2x\). The left side oscillates between -3 and 3, while the right side is a linear function. There is one intersection point within the given range \(0 \leq x \leq 4\pi\).

(i) The function \(f(x) = 2 - 3 \cos x\) has a range determined by the range of \(\cos x\), which is \([-1, 1]\). Therefore, the range of \(f(x)\) is \(2 - 3(-1) \leq f(x) \leq 2 - 3(1)\), which simplifies to \(-1 \leq f(x) \leq 5\).

(ii) The graph of \(y = f(x)\) is a cosine wave starting at \((0, -1)\), peaking at \((\pi, 5)\), and ending at \((2\pi, -1)\). It is symmetrical about \(x = \pi\).

(iii) For \(g(x) = 2 - 3 \cos x\) to have an inverse, it must be one-to-one. The largest interval for \(x\) where \(\cos x\) is one-to-one is \([0, \pi]\). Thus, the largest value of \(p\) is \(\pi\).

(iv) To find \(g^{-1}(x)\), set \(y = 2 - 3 \cos x\) and solve for \(x\):

\(y = 2 - 3 \cos x\)

\(\cos x = \frac{2-y}{3}\)

\(x = \cos^{-1} \left( \frac{2-y}{3} \right)\)

Thus, \(g^{-1}(x) = \cos^{-1} \left( \frac{2-x}{3} \right)\).

(i) Start with the equation \(f(x) + 4 = 0\), which simplifies to \(3 \tan\left(\frac{1}{2}x\right) - 2 + 4 = 0\) or \(3 \tan\left(\frac{1}{2}x\right) = -2\). This gives \(\tan\left(\frac{1}{2}x\right) = -\frac{2}{3}\). Solving for \(\frac{1}{2}x\), we find \(\frac{1}{2}x = \tan^{-1}\left(-\frac{2}{3}\right) \approx -0.588\). Therefore, \(x = 2(-0.588) = -1.2\).

(ii) To find \(f^{-1}(x)\), start with \(y = 3 \tan\left(\frac{1}{2}x\right) - 2\). Rearrange to isolate \(\tan\left(\frac{1}{2}x\right)\): \(y + 2 = 3 \tan\left(\frac{1}{2}x\right)\), so \(\tan\left(\frac{1}{2}x\right) = \frac{y+2}{3}\). Taking the inverse tangent gives \(\frac{1}{2}x = \tan^{-1}\left(\frac{y+2}{3}\right)\), thus \(x = 2 \tan^{-1}\left(\frac{y+2}{3}\right)\). Therefore, \(f^{-1}(x) = 2 \tan^{-1}\left(\frac{x+2}{3}\right)\). The domain of \(f^{-1}\) is \(-5, 1\) based on the range of \(f\).

(iii) The sketch should show a tan graph through the first, third, and fourth quadrants, and an inverse tan graph through the first, second, and third quadrants. The two curves should be symmetrical about \(y = x\) and approximately in the correct domain and range.

(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 \(x = \frac{1}{2} \sin^{-1} \left( \frac{5-x}{2} \right)\).