(i) We have two equations from the given conditions:

\(a \sin\left(\frac{1}{2}\pi\right) + b = 2\)

\(a \cdot 1 + b = 2\)

\(a + b = 2\)

\(a \sin\left(\frac{3}{2}\pi\right) + b = -8\)

\(a \cdot (-1) + b = -8\)

\(-a + b = -8\)

Solving these equations:

\(a + b = 2\)

\(-a + b = -8\)

Add the equations:

\(2b = -6\)

\(b = -3\)

Substitute \(b = -3\) into \(a + b = 2\):

\(a - 3 = 2\)

\(a = 5\)

(ii) To find \(x\) for which \(f(x) = 0\):

\(5 \sin x - 3 = 0\)

\(5 \sin x = 3\)

\(\sin x = \frac{3}{5}\)

\(x = \sin^{-1}\left(\frac{3}{5}\right) \approx 0.64\)

Since \(\sin x\) is positive in the first and second quadrants:

\(x = \pi - 0.64 \approx 2.50\)

(iii) The graph of \(y = 5 \sin x - 3\) is a sine wave with amplitude 5, shifted down by 3 units. It starts at \(y = -3\) when \(x = 0\), reaches a maximum of \(y = 2\) at \(x = \frac{1}{2}\pi\), and a minimum of \(y = -8\) at \(x = \frac{3}{2}\pi\).