Answer: (a)(i) \(1+\frac57x+\frac{10}{49}x^2\); (a)(ii) \(n=12\); (b) \(k=-4\).

Use the binomial theorem carefully, keeping track of the power of each factor in the general term.

(a)(i) Using the binomial expansion,

So the first three terms are

\(1+\frac57x+\frac{10}{49}x^2.\)

(a)(ii) For the first two terms,

\((1+x)^n=1+nx+\cdots\)

and

\(\left(1+\frac{x}{7}\right)^5=1+\frac57x+\cdots.\)

Thus

\(7(1+x)^n\left(1+\frac{x}{7}\right)^5 =7\left(1+\left(n+\frac57\right)x+\cdots\right).\)

So the coefficient of \(x\) is

\(7n+5.\)

Given that the first two terms are \(7+89x\),

\(7n+5=89.\)

Hence

\(n=12.\)

(b) In the expansion of \((k-2x)^8\), the coefficient of \(x^r\) is

\(\binom8r k^{8-r}(-2)^r.\)

The coefficient of \(x^4\) is

\(\binom84 k^4(-2)^4=70k^4\cdot16=1120k^4.\)

The coefficient of \(x^2\) is

\(\binom82 k^6(-2)^2=28k^6\cdot4=112k^6.\)

The given ratio is therefore

\(\frac{1120k^4}{112k^6}=\frac58.\)

So

\(\frac{10}{k^2}=\frac58.\)

Hence

\(k^2=16.\)

Therefore \(k=4\) or \(k=-4\).

The coefficient of \(x\) is

\(\binom81k^7(-2)=-16k^7.\)

This is positive, so \(k\) must be negative. Hence

\(k=-4.\)