(i) To find the first three terms in the expansion of \((2-x)^6\), we use the binomial theorem:

\((2-x)^6 = \sum_{r=0}^{6} \binom{6}{r} (2)^{6-r} (-x)^r\)

The first three terms are:

\(\binom{6}{0} (2)^6 (-x)^0 = 64\)

\(\binom{6}{1} (2)^5 (-x)^1 = -192x\)

\(\binom{6}{2} (2)^4 (-x)^2 = 240x^2\)

Thus, the first three terms are \(64 - 192x + 240x^2\).

(ii) For the expansion of \((1+kx)(2-x)^6\), we need the coefficient of \(x^2\) to be zero.

The expansion of \((2-x)^6\) gives the \(x^2\) term as \(240x^2\).

When multiplying by \(1+kx\), the \(x^2\) term comes from:

\(k \times (-192x) \times x = -192kx^2\)

Thus, the coefficient of \(x^2\) is \(240 - 192k\).

Setting this to zero gives:

\(240 - 192k = 0\)

\(192k = 240\)

\(k = \frac{240}{192} = \frac{5}{4} = 1.25\)