(a) To expand \((1 + 3x)^6\) up to the term in \(x^2\), we use the binomial theorem:

\((1 + 3x)^6 = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (3x)^k\)

Calculating the first three terms:

\(\text{Term in } x^0: \binom{6}{0} (1)^6 (3x)^0 = 1\)

\(\text{Term in } x^1: \binom{6}{1} (1)^5 (3x)^1 = 18x\)

\(\text{Term in } x^2: \binom{6}{2} (1)^4 (3x)^2 = 135x^2\)

Thus, the expansion up to \(x^2\) is \(1 + 18x + 135x^2\).

(b) To find the coefficient of \(x^2\) in \((1 - 7x + x^2)(1 + 3x)^6\), we consider the relevant terms:

1. \(x^2\) term from \((1 + 3x)^6\) is \(135x^2\).

2. \(x^1\) term from \((1 + 3x)^6\) is \(18x\), and multiplying by \(-7x\) from \((1 - 7x + x^2)\) gives \(-126x^2\).

3. \(x^0\) term from \((1 + 3x)^6\) is \(1\), and multiplying by \(x^2\) from \((1 - 7x + x^2)\) gives \(x^2\).

Adding these contributions: \(135x^2 - 126x^2 + x^2 = 10x^2\).

Therefore, the coefficient of \(x^2\) is 10.

(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 \(a\):

\(42 - 12 = 10a\)

\(30 = 10a\)

\(a = 3\)

(i) The binomial expansion of \((a-x)^5\) is given by:

\(a^5 - 5a^4x + 10a^3x^2 - 10a^2x^3 + \ldots\)

Thus, the first four terms in ascending powers of \(x\) are \(a^5 - 5a^4x + 10a^3x^2 - 10a^2x^3\).

(ii) To find the coefficient of \(x^3\) in the expansion of \((1-ax)(a-x)^5\), consider the terms that contribute to \(x^3\):

The term from \((a-x)^5\) is \(-10a^2x^3\).

The term from \((1-ax)\) is \(-ax\), which when multiplied by the \(x^2\) term \(10a^3x^2\) from \((a-x)^5\), gives \(-10a^4x^3\).

Thus, the coefficient of \(x^3\) is:

\(-10a^4 - 10a^2 = -200\)

Simplifying, we get:

\(-10(a^4 + a^2) = -200\)

\(a^4 + a^2 = 20\)

Let \(y = a^2\), then:

\(y^2 + y - 20 = 0\)

Solving this quadratic equation using the quadratic formula:

\(y = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2}\)

\(y = 4\) or \(y = -5\)

Since \(y = a^2\), \(y\) must be non-negative, so \(y = 4\).

Thus, \(a^2 = 4\), giving \(a = \pm 2\).

(i) (a) The expansion of \((1-x)^6\) using the binomial theorem is:

\((1-x)^6 = 1 - 6x + 15x^2 + \ldots\)

(b) The expansion of \((1+2x)^6\) using the binomial theorem is:

\((1+2x)^6 = 1 + 12x + 60x^2 + \ldots\)

(ii) To find the coefficient of \(x^2\) in \([(1-x)(1+2x)]^6\), we consider the product of the expansions:

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

\((1+2x)^6 = 1 + 12x + 60x^2\)

The coefficient of \(x^2\) is obtained by considering the terms:

\(1 \cdot 60x^2 = 60x^2\)

\(-6x \cdot 12x = -72x^2\)

\(15x^2 \cdot 1 = 15x^2\)

Adding these gives the coefficient of \(x^2\):

\(60 - 72 + 15 = 3\)

(i) To find the first three terms in the expansion of \((1 + x)^5\), we use the binomial theorem:

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

The first three terms are:

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

\(\binom{5}{1} x^1 = 5x\)

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

Thus, the first three terms are \(1 + 5x + 10x^2\).

(ii) For the expansion of \((1 + (px + x^2))^5\), we replace \(x\) by \((px + x^2)\) in the expansion of \((1 + x)^5\):

\((1 + (px + x^2))^5 = 1 + 5(px + x^2) + 10(px + x^2)^2\)

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

From \(5(px + x^2)\), the \(x^2\) term is \(5x^2\).

From \(10(px + x^2)^2\), the \(x^2\) term is \(10p^2x^2\).

Thus, the coefficient of \(x^2\) is \(5 + 10p^2\).

We are given that this coefficient is 95:

\(5 + 10p^2 = 95\)

\(10p^2 = 90\)

\(p^2 = 9\)

\(p = 3\) (since \(p\) is positive)