Answer: (a) \(2x-y=\frac{\pi}{4}-4\). (b) \(PQ\approx3.59\).

From

\(2y=\tan 2x+7,\)

we have

\(y=\frac12\tan 2x+\frac72.\)

Differentiate:

\(\frac{dy}{dx}=\operatorname{sec}^2 2x.\)

At \(x=\frac{\pi}{8}\),

\(2x=\frac{\pi}{4},\qquad \operatorname{sec}^2\frac{\pi}{4}=2.\)

Also

\(y=\frac12\tan\frac{\pi}{4}+\frac72=4.\)

The tangent has gradient \(2\) and passes through \(\left(\frac{\pi}{8},4\right)\), so

\(y-4=2\left(x-\frac{\pi}{8}\right).\)

Hence

\(y=2x-\frac{\pi}{4}+4,\)

or

\(\boxed{2x-y=\frac{\pi}{4}-4}.\)

For \(P\), put \(y=0\):

\(0=2x-\frac{\pi}{4}+4,\)

so

\(x=\frac{\pi}{8}-2.\)

Thus \(P=\left(\frac{\pi}{8}-2,0\right)\).

For \(Q\), put \(x=0\):

\(y=4-\frac{\pi}{4}.\)

Thus \(Q=\left(0,4-\frac{\pi}{4}\right)\).

Therefore

\(PQ=\sqrt{\left(\frac{\pi}{8}-2\right)^2+\left(4-\frac{\pi}{4}\right)^2}.\)

Numerically,

\(\boxed{PQ\approx3.59}.\)