Answer: (a) \(5\); (b) \(480^\circ\).

For a trigonometric graph, identify the amplitude, midline and period first, then use these to locate the key points of the sketch.

(a) The function is

\(g(x)=5\sin\left(\frac{3x}{4}\right)-2\).

The amplitude is the coefficient of the sine function, so the amplitude is \(5\).

(b) For \(\sin(kx)\), the period in degrees is \(\frac{360^\circ}{k}\).

Here \(k=\frac34\), so the period is

\(\frac{360^\circ}{3/4}=480^\circ\).

(c) The graph has midline \(y=-2\), amplitude \(5\), maximum value \(3\), and minimum value \(-7\).

It passes through \((0,-2)\). In the interval \(-180^\circ\leq x\leq180^\circ\), it has a minimum at \((-120^\circ,-7)\) and a maximum at \((120^\circ,3)\).

It crosses the \(x\)-axis once between \(0^\circ\) and \(60^\circ\). The curve starts in the third quadrant, rises through \((0,-2)\), reaches its maximum in the first quadrant, and then ends with \(1\lt y\lt 2\) at \(x=180^\circ\).

Therefore the result matches the required answer.