Answer: (i) \(256x^8-64x^6+7x^4\). (ii) \(135\).

(i) Use the binomial theorem.

The first term is

\((2x)^8=256x^8.\)

The second term is

\({8\choose1}(2x)^7\left(-\frac{1}{16x}\right) = 8\cdot128x^7\left(-\frac{1}{16x}\right) = -64x^6.\)

The third term is

\({8\choose2}(2x)^6\left(-\frac{1}{16x}\right)^2.\)

So

\({8\choose2}(2x)^6\left(\frac{1}{256x^2}\right) = 28\cdot64x^6\cdot\frac{1}{256x^2} = 7x^4.\)

Hence the first three terms are

\(256x^8-64x^6+7x^4.\)

(ii) Expand the second factor:

\(\left(\frac{1}{x^2}+1\right)^2 = \frac{1}{x^4}+\frac{2}{x^2}+1.\)

To obtain an \(x^4\) term, use:

\(256x^8\cdot\frac{1}{x^4}, \qquad -64x^6\cdot\frac{2}{x^2}, \qquad 7x^4\cdot1.\)

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

\(256-128+7=135.\)