(i) To find the first three terms of \((1 - 2x)^5\), use the binomial expansion formula:

\((1 - 2x)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-2x)^k\)

The first three terms are:

\(\binom{5}{0}(1)^5(-2x)^0 = 1\)

\(\binom{5}{1}(1)^4(-2x)^1 = -10x\)

\(\binom{5}{2}(1)^3(-2x)^2 = 40x^2\)

Thus, the first three terms are \(1 - 10x + 40x^2\).

(ii) Expand \((1 + ax + 2x^2)(1 - 10x + 40x^2)\) and find the coefficient of \(x^2\):

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

\(1 \cdot 40x^2 = 40x^2\)

\(ax \cdot (-10x) = -10ax^2\)

\(2x^2 \cdot 1 = 2x^2\)

Combine these to get \(40x^2 - 10ax^2 + 2x^2 = (42 - 10a)x^2\).

Set the coefficient equal to 12:

\(42 - 10a = 12\)

Solve for \(a\):

\(42 - 12 = 10a\)

\(30 = 10a\)

\(a = 3\)