(a) To find the first three terms in the expansion of \((1 + ax)^6\), we use the binomial theorem:

\((1 + ax)^6 = 1 + \binom{6}{1}(ax) + \binom{6}{2}(ax)^2 + \ldots\)

Calculating the first three terms:

\(1 + 6ax + 15a^2x^2\)

(b) To find the coefficient of \(x^2\) in the expansion of \((1 - 3x)(1 + ax)^6\), we first find the relevant terms from each factor:

The expansion of \((1 + ax)^6\) gives the terms \(1, 6ax, 15a^2x^2\).

We need the coefficient of \(x^2\) from the product:

\((1)(15a^2x^2) + (-3x)(6ax) = 15a^2x^2 - 18ax^2\)

Setting the coefficient equal to \(-3\):

\(15a^2 - 18a = -3\)

Rearranging gives:

\(15a^2 - 18a + 3 = 0\)

Factoring the quadratic:

\(3(5a^2 - 6a + 1) = 0\)

\((3)(a-1)(5a-1) = 0\)

Thus, \(a = 1\) or \(a = \frac{1}{5}\).