Answer: (a) \(n=5,\ p=2,\ q=1080\); (b) \(\dfrac{80}{3}\).

We start with the main method. Use the binomial theorem to identify only the terms needed, rather than expanding the whole expression.

The constant term in \((3+px)^n\) is \(3^n\).

So

\(3^n=243=3^5,\)

giving

\(n=5.\)

The coefficient of \(x\) is

\(\binom51 3^4p=5\cdot81p.\)

Since this coefficient is \(810\),

\(405p=810,\)

so

\(p=2.\)

The coefficient of \(x^2\) is

\(\binom52 3^3p^2=10\cdot27\cdot 2^2=1080.\)

Thus \(q=1080\).

For part (b), the general term in

\(\left(2y-\frac{1}{3y^2}\right)^6\)

is

\(\binom6r(2y)^{6-r}\left(-\frac{1}{3y^2}\right)^r.\)

The power of \(y\) is

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

For the term independent of \(y\),

\(6-3r=0,\)

so \(r=2\).

The required term is

\(\binom62(2y)^4\left(-\frac{1}{3y^2}\right)^2 =15\cdot16\cdot\frac19=\frac{80}{3}.\)

This completes the solution and gives the required result.