Answer: (i) \(-\dfrac{5}{x}+\dfrac{5}{8x^3}-\dfrac{1}{32x^5}\). (ii) The coefficient is \(0\).

(i) The general term in the expansion of \(\left(2x-\dfrac{1}{2x}\right)^5\) is

\(\binom{5}{r}(2x)^{5-r}\left(-\frac{1}{2x}\right)^r.\)

The first three terms are given as

\(32x^5-40x^3+20x.\)

The remaining terms are obtained from \(r=3,4,5\).

For \(r=3\),

\(\binom{5}{3}(2x)^2\left(-\frac{1}{2x}\right)^3=-\frac{5}{x}.\)

For \(r=4\),

\(\binom{5}{4}(2x)\left(-\frac{1}{2x}\right)^4=\frac{5}{8x^3}.\)

For \(r=5\),

\(\left(-\frac{1}{2x}\right)^5=-\frac{1}{32x^5}.\)

So the remaining three terms are

\(-\frac{5}{x}+\frac{5}{8x^3}-\frac{1}{32x^5}.\)

(ii) In \((1+4x^2)\left(2x-\dfrac{1}{2x}\right)^5\), the terms that can contribute to the coefficient of \(x\) are:

from multiplying by \(1\), the term \(20x\), and from multiplying by \(4x^2\), the term \(-\dfrac{5}{x}\).

Thus the coefficient of \(x\) is

\(20+4(-5)=20-20=0.\)