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

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

The first three terms are:

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

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

\(\binom{5}{2} (2)^3 (ax)^2 = 80a^2x^2\).

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

(ii) We need to find the coefficient of \(x^2\) in the expansion of \((1 + 2x)(2 + ax)^5\).

First, expand \((2 + ax)^5\) to get the terms involving \(x^2\):

\(32 + 80ax + 80a^2x^2\).

Now, multiply by \((1 + 2x)\):

\((32 + 80ax + 80a^2x^2)(1 + 2x)\).

The terms contributing to \(x^2\) are:

\(32 \cdot 2x = 64x\),

\(80ax \cdot 1 = 80ax\),

\(80a^2x^2 \cdot 1 = 80a^2x^2\).

The coefficient of \(x^2\) is \(80a^2 + 160a\).

Set this equal to 240:

\(80a^2 + 160a = 240\).

Divide by 80:

\(a^2 + 2a = 3\).

Rearrange to form a quadratic equation:

\(a^2 + 2a - 3 = 0\).

Factorize:

\((a - 1)(a + 3) = 0\).

Thus, \(a = 1\) or \(a = -3\).