Answer: (a) \(\dfrac{105}{8}\); (b)(i) \(k=144\); (b)(ii) \(288-144\sqrt2\).

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

For part (a), the general term in the expansion is

\({}^{10}C_r(x^2)^{10-r}\left(-\frac{1}{2x^3}\right)^r.\)

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

\(2(10-r)-3r=20-5r.\)

For a term independent of \(x\), set \(20-5r=0\). This gives \(r=4\).

The required term is therefore

\({}^{10}C_4(x^2)^6\left(-\frac{1}{2x^3}\right)^4=\frac{{}^{10}C_4}{16}=\frac{210}{16}=\frac{105}{8}.\)

For part (b), expand both fourth powers:

\((1+2\sqrt2)^4\)

\(=1+8\sqrt2+48+64\sqrt2+64\)

\(=113+72\sqrt2.\)

Similarly,

\((1-2\sqrt2)^4=113-72\sqrt2.\)

Therefore

\((1+2\sqrt2)^4-(1-2\sqrt2)^4=144\sqrt2,\)

so \(k=144\).

Now

\(\frac{144\sqrt2}{1+\sqrt2}\cdot\frac{1-\sqrt2}{1-\sqrt2} =\frac{144\sqrt2-288}{-1}.\)

Therefore the value is \(288-144\sqrt2.\)

This completes the solution and gives the required result.