Answer: (a) \(-540\). (b) \(n=20\).

(a) The general term in

\(\left(3x-\frac1x\right)^6\)

is

\(\binom6r(3x)^{6-r}\left(-\frac1x\right)^r.\)

This is

\(\binom6r3^{6-r}(-1)^r x^{6-2r}.\)

For the term independent of \(x\),

\(6-2r=0,\)

so \(r=3\). Therefore the independent term is

\(\binom63 3^3(-1)^3=20\cdot27\cdot(-1)=-540.\)

So the term is

\(\boxed{-540}.\)

(b) The coefficient of \(x^4\) is

\(\binom n4\left(\frac12\right)^4.\)

The coefficient of \(x^6\) is

\(\binom n6\left(\frac12\right)^6.\)

Given that the coefficient of \(x^4\) is half the coefficient of \(x^6\),

\(\binom n4\frac{1}{16}=\frac12\binom n6\frac{1}{64}.\)

So

\(8\binom n4=\binom n6.\)

Using

\(\frac{\binom n6}{\binom n4}=\frac{(n-4)(n-5)}{30},\)

we get

\(\frac{(n-4)(n-5)}{30}=8.\)

Hence

\((n-4)(n-5)=240.\)

So

\(n^2-9n-220=0.\)

Factorise:

\((n-20)(n+11)=0.\)

Since \(n\) is positive,

\(\boxed{n=20}.\)