(i) To find the first three terms in the expansion of \((2+u)^5\), we use the binomial theorem:

\((2+u)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} u^k\).

The first three terms are:

\(\binom{5}{0} 2^5 u^0 = 32\),

\(\binom{5}{1} 2^4 u^1 = 80u\),

\(\binom{5}{2} 2^3 u^2 = 80u^2\).

Thus, the first three terms are \(32 + 80u + 80u^2\).

(ii) Substitute \(u = x + x^2\) into the expansion:

\(80(x + x^2) + 80(x + x^2)^2\).

Calculate the coefficient of \(x^2\):

From \(80(x + x^2)\), the coefficient of \(x^2\) is \(80\).

From \(80(x + x^2)^2\), expand \((x + x^2)^2 = x^2 + 2x^3 + x^4\), the coefficient of \(x^2\) is \(80\).

Adding these gives the total coefficient of \(x^2\) as \(80 + 80 = 160\).