Answer: \(k=2\); \((1+3x)^6=1+18x+135x^2+\cdots\); \(a=-2,\ b=469\).

(a) In

\(\left(2k-\frac{x}{k}\right)^5,\)

the term in \(x^2\) is obtained by choosing \(\left(-\frac{x}{k}\right)^2\) and \((2k)^3\):

\(\binom52(2k)^3\left(-\frac{x}{k}\right)^2.\)

The coefficient of \(x^2\) is therefore

\(10\cdot8k^3\cdot\frac1{k^2}=80k.\)

Since this coefficient is \(160\),

\(80k=160,\)

so

\(k=2.\)

(b)(i) The first three terms of \((1+3x)^6\) are

\(1+\binom61(3x)+\binom62(3x)^2.\)

Thus

\((1+3x)^6=1+18x+135x^2+\cdots.\)

(b)(ii) Now

\((a+x)^2=a^2+2ax+x^2.\)

The constant term of \((1+3x)^6(a+x)^2\) is \(a^2\). Since the constant term is \(4\),

\(a^2=4,\)

so \(a=2\) or \(a=-2\).

The coefficient of \(x\) is

\(18a^2+2a.\)

This is \(68\), so

\(18(4)+2a=68.\)

Hence

\(72+2a=68,\)

and therefore

\(a=-2.\)

The coefficient of \(x^2\) is

\(135a^2+18(2a)+1.\)

Substitute \(a=-2\):

\(b=135(4)+36(-2)+1=540-72+1=469.\)