Answer: \(p=2\), \(q=7\), \(r=-7\); no constant term; \(n=16\).

In the expansion of \(\left(x^2-\frac1{px}\right)^8\), the general term is

\(\binom8s(x^2)^{8-s}\left(-\frac1{px}\right)^s.\)

This simplifies to

\(\binom8s(-1)^s p^{-s}x^{16-3s}.\)

For the \(x^{13}\) term, \(16-3s=13\), so \(s=1\). Its coefficient is

\(-\frac8p.\)

Given that this coefficient is \(-4\),

\(-\frac8p=-4,\)

so

\(\boxed{p=2}.\)

For the \(x^{10}\) term, \(s=2\). The coefficient is

\(\binom82\frac1{p^2}=\frac{28}{4}=7,\)

so \(\boxed{q=7}\).

For the \(x^7\) term, \(s=3\). The coefficient is

\(-\binom83\frac1{p^3}=-\frac{56}{8}=-7,\)

so \(\boxed{r=-7}\).

A term independent of \(x\) would require

\(16-3s=0.\)

This gives \(s=\frac{16}{3}\), which is not an integer, so no such term exists.

For part (b), the \(x\) term comes from the term containing \((\sqrt{x})^2=x\). Thus use \(s=2\):

\(\binom n2\left(-\frac12\right)^2=30.\)

So

\(\frac{n(n-1)}2\cdot\frac14=30.\)

Hence

\(n(n-1)=240.\)

Therefore

\(n^2-n-240=0.\)

Factorising,

\((n-16)(n+15)=0.\)

Since \(n\) is positive,

\(\boxed{n=16}.\)