Answer: period \(1080^\circ\); the sketch is symmetric about the \(y\)-axis, has \(y\)-intercept \(1\), and \(x\)-intercepts at \(-180^\circ\) and \(180^\circ\).

(a) For \(y=2\cos\frac{x}{3}-1\), the angle inside the cosine is \(\frac{x}{3}\).

A full cycle occurs when \(\frac{x}{3}\) increases by \(360^\circ\), so \(x\) increases by

\(3\times360^\circ=1080^\circ.\)

Therefore the period is

\(1080^\circ.\)

(b) The graph has midline \(y=-1\) and amplitude \(2\), so its maximum value is \(1\) and its minimum value is \(-3\).

At \(x=0\),

\(y=2\cos0^\circ-1=1,\)

so the graph has \(y\)-intercept \(1\).

For the \(x\)-intercepts, set \(y=0\):

\(2\cos\frac{x}{3}-1=0.\)

So

\(\cos\frac{x}{3}=\frac12.\)

In the interval \(-360^\circ\leq x\leq360^\circ\), this gives

\(x=-180^\circ\quad\text{or}\quad x=180^\circ.\)

The sketch should be a smooth cosine curve, symmetric about the \(y\)-axis, passing through \((-180^\circ,0)\), \((0,1)\) and \((180^\circ,0)\).