Answer: period \(4\pi\), \(y\)-intercept \(1\), asymptotes \(x=-2\pi\) and \(x=2\pi\).

(a) The basic period of \(\tan x\) is \(\pi\). For

\(\tan\left(\frac{x}{4}\right),\)

the input \(\frac{x}{4}\) must increase by \(\pi\) for one full period. Therefore \(x\) must increase by

\(4\pi.\)

So the period is

\(4\pi.\)

(b) The graph is

\(y=5\tan\left(\frac{x}{4}\right)+1.\)

The \(y\)-intercept is found by putting \(x=0\):

\(y=5\tan0+1=1.\)

So the \(y\)-intercept is \((0,1)\).

Vertical asymptotes occur when

\(\frac{x}{4}=\frac{\pi}{2}+n\pi.\)

Hence

\(x=2\pi+4n\pi.\)

For \(-2\pi\leqslant x\leqslant4\pi\), the asymptotes are

\(x=-2\pi\quad\text{and}\quad x=2\pi.\)

The sketch should show a tangent curve shifted up by \(1\), passing through \((0,1)\), with these two vertical asymptotes.