(i) The binomial expansion of \((k + x)^8\) is given by:

\(\sum_{r=0}^{8} \binom{8}{r} k^{8-r} x^r\)

The first four terms are:

\(\binom{8}{0} k^8 x^0 = k^8\)

\(\binom{8}{1} k^7 x^1 = 8k^7x\)

\(\binom{8}{2} k^6 x^2 = 28k^6x^2\)

\(\binom{8}{3} k^5 x^3 = 56k^5x^3\)

Thus, the first four terms are \(k^8 + 8k^7x + 28k^6x^2 + 56k^5x^3\).

(ii) Given that the coefficients of \(x^2\) and \(x^3\) are equal, we have:

\(28k^6 = 56k^5\)

Dividing both sides by \(28k^5\), we get:

\(k = 2\)