Answer: (a) \(n=16\), \(p=30\), \(q=-70\); (b) \(\frac{20}{27}\).

Use the binomial theorem carefully, keeping track of the power of each factor in the general term.

(a) The expansion of \(\left(1-\frac{x}{2}\right)^n\) begins

The coefficient of \(x\) is \(-\frac n2\). This is given as \(-8\), so

\(-\frac n2=-8\).

Hence \(n=16\).

The coefficient of \(x^2\) is

\(\frac{n(n-1)}{2!}\cdot\frac14=\frac{n(n-1)}8\).

With \(n=16\),

\(p=\frac{16\cdot15}{8}=30\).

The coefficient of \(x^3\) is

\(\frac{n(n-1)(n-2)}{3!}\left(-\frac18\right)=-\frac{n(n-1)(n-2)}{48}\).

With \(n=16\),

\(q=-\frac{16\cdot15\cdot14}{48}=-70\).

Therefore \(n=16\), \(p=30\), and \(q=-70\).

(b) The general term in \(\left(\frac{2}{x^2}+\frac{x}{3}\right)^6\) is

\(\binom6r\left(\frac{2}{x^2}\right)^{6-r}\left(\frac{x}{3}\right)^r\).

The power of \(x\) in this term is

\(-2(6-r)+r=-12+3r\).

For the term independent of \(x\), set this equal to 0:

\(-12+3r=0\).

So \(r=4\).

The required term is

\(\binom64\left(\frac{2}{x^2}\right)^2\left(\frac{x}{3}\right)^4\).

This is

\(15\cdot\frac4{x^4}\cdot\frac{x^4}{81}=\frac{60}{81}=\frac{20}{27}\).