Answer: (a) \(3\sqrt3\); (b) \(\left(-10+\sqrt{17},-\dfrac{10+\sqrt{17}}{83}\right)\) and \(\left(-10-\sqrt{17},-\dfrac{10-\sqrt{17}}{83}\right)\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

For part (a), the curve cuts the \(y\)-axis where \(x=0\).

Substitute \(x=0\) into

\(9(x-1)^2+4(y-3)^2=36.\)

This gives

\(9+4(y-3)^2=36.\)

So

\(4(y-3)^2=27\), and hence

\(y=3\pm\sqrt{\frac{27}{4}}=3\pm\frac{3\sqrt3}{2}.\)

The distance between the two points on the \(y\)-axis is therefore

\(\left(3+\frac{3\sqrt3}{2}\right)-\left(3-\frac{3\sqrt3}{2}\right)=3\sqrt3.\)

For part (b), substitute \(y=\frac1x\) into the other curve:

\(2x^2+83x\left(\frac1x\right)=x^3\left(\frac1x\right)-20x.\)

So

\(2x^2+83=x^2-20x.\)

Hence

\(x^2+20x+83=0.\)

Using the quadratic formula,

\(x=\frac{-20\pm\sqrt{20^2-4(1)(83)}}{2}.\)

Thus

\(x=\frac{-20\pm\sqrt{68}}{2}=-10\pm\sqrt{17}.\)

Now \(y=\frac1x\). For \(x=-10+\sqrt{17}\),

\(y=\frac{1}{-10+\sqrt{17}}\cdot\frac{-10-\sqrt{17}}{-10-\sqrt{17}} =\frac{-10-\sqrt{17}}{100-17} =-\frac{10+\sqrt{17}}{83}.\)

For \(x=-10-\sqrt{17}\),

\(y=\frac{1}{-10-\sqrt{17}}\cdot\frac{-10+\sqrt{17}}{-10+\sqrt{17}} =\frac{-10+\sqrt{17}}{100-17} =-\frac{10-\sqrt{17}}{83}.\)

Therefore the two points are

\(\left(-10+\sqrt{17},-\frac{10+\sqrt{17}}{83}\right)\)

and

\(\left(-10-\sqrt{17},-\frac{10-\sqrt{17}}{83}\right).\)

This completes the solution and gives the required result.