(i) To find the first three terms of \((2 + 3x)^6\), we use the binomial expansion formula:

\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

For \((2 + 3x)^6\), \(a = 2\), \(b = 3x\), and \(n = 6\).

The first three terms are:

1st term: \(\binom{6}{0} 2^6 (3x)^0 = 64\)

2nd term: \(\binom{6}{1} 2^5 (3x)^1 = 192 \times 3x = 576x\)

3rd term: \(\binom{6}{2} 2^4 (3x)^2 = 15 \times 16 \times 9x^2 = 2160x^2\)

Thus, the first three terms are \(64 + 576x + 2160x^2\).

(ii) In the expansion of \((1 + ax)(2 + 3x)^6\), the coefficient of \(x^2\) is zero.

The term \(x^2\) comes from two sources:

1. The \(x^2\) term from \((2 + 3x)^6\), which is \(2160x^2\).

2. The product of the \(x\) term from \((1 + ax)\) and the \(x\) term from \((2 + 3x)^6\), which is \(576ax^2\).

Setting the sum of these coefficients to zero:

\(576a + 2160 = 0\)

Solving for \(a\):

\(576a = -2160\)

\(a = -\frac{2160}{576} = -\frac{15}{4}\)

Thus, \(a = -\frac{15}{4}\).

(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 \(a = -3\).

(i) To find the first three terms of \((1 - px)^6\), use the binomial expansion formula:

\((1 - px)^6 = \sum_{k=0}^{6} \binom{6}{k} (-px)^k\).

The first three terms are:

\(\binom{6}{0}(1)^6 = 1\)

\(\binom{6}{1}(-px) = -6px\)

\(\binom{6}{2}(-px)^2 = 15p^2x^2\)

Thus, the first three terms are \(1 - 6px + 15p^2x^2\).

(ii) The expansion of \((1 - x)(1 - px)^6\) involves multiplying \((1 - x)\) by the expansion of \((1 - px)^6\). The coefficient of \(x^2\) is given by:

\(-6p \times (-x) + 15p^2x^2 \times 1 = 0\)

\(15p^2 - 6p = 0\)

Factor the quadratic equation:

\(3p(5p + 2) = 0\)

Thus, \(p = 0\) or \(p = -\frac{2}{5}\).

Since \(p\) is non-zero, \(p = -\frac{2}{5}\).

(i) To expand \((2x - x^2)^6\), use the binomial theorem:

\((a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k\)

Here, \(a = 2x\), \(b = x^2\), and \(n = 6\).

The first three terms are:

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

\(\binom{6}{1} (2x)^5 (x^2)^1 = -192x^7\)

\(\binom{6}{2} (2x)^4 (x^2)^2 = 240x^8\)

Thus, the first three terms are \(64x^6 - 192x^7 + 240x^8\).

(ii) To find the coefficient of \(x^8\) in \((2 + x)(2x - x^2)^6\), consider the terms that contribute to \(x^8\):

1. \((2)(240x^8) = 480x^8\)

2. \((x)(-192x^7) = -192x^8\)

The coefficient of \(x^8\) is \(480 - 192 = 288\).

First, expand \((1 + ax)^6\) using the binomial theorem:

The coefficient of \(x\) in \((1 + ax)^6\) is \(6ax\).

The coefficient of \(x^2\) in \((1 + ax)^6\) is \(15a^2x^2\).

Now, consider the expansion of \((1 - 2x)^2 = 1 - 4x + 4x^2\).

Multiply by \((1 + ax)^6\):

For the \(x\) term: \(6a - 4 = -1\), solving gives \(a = \frac{1}{2}\).

For the \(x^2\) term: \(15a^2 - 24a + 4 = b\), substituting \(a = \frac{1}{2}\) gives \(b = -1\).