We start by rewriting \(\frac{16}{(2+x)^2}\) as \(16(2+x)^{-2}\).

Using the binomial expansion for \((1 + u)^n\), where \(u = \frac{x}{2}\) and \(n = -2\), we have:

\((1 + \frac{x}{2})^{-2} = 1 - 2\frac{x}{2} + \frac{(-2)(-3)}{2!} \left(\frac{x}{2}\right)^2 + \cdots\)

Simplifying, we get:

\(1 - x + \frac{3}{2}x^2 + \cdots\)

Now, multiply by 16:

\(16(1 - x + \frac{3}{2}x^2) = 16 - 16x + 24x^2\)

Thus, the expansion up to \(x^2\) is \(4 - 4x + 3x^2\).

We start by expanding \(\sqrt[3]{1 - 6x}\) using the binomial series expansion for \((1 + u)^n\), where \(n = \frac{1}{3}\) and \(u = -6x\).

The binomial expansion is given by:

\((1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots\)

Substituting \(n = \frac{1}{3}\) and \(u = -6x\), we have:

\(1 + \frac{1}{3}(-6x) + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!}(-6x)^2 + \frac{\frac{1}{3}(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}(-6x)^3\)

Calculating each term:

First term: \(1\)

Second term: \(\frac{1}{3}(-6x) = -2x\)

Third term: \(\frac{\frac{1}{3}(-\frac{2}{3})}{2}(-6x)^2 = \frac{1}{3} \times -\frac{2}{3} \times 18x^2 = -4x^2\)

Fourth term: \(\frac{\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})}{6}(-6x)^3 = \frac{1}{3} \times -\frac{2}{3} \times -\frac{5}{3} \times -216x^3 = -\frac{40}{3}x^3\)

Thus, the expansion up to \(x^3\) is:

\(1 - 2x - 4x^2 - \frac{40}{3}x^3\)

To expand \((1 + 2x)^{-3}\), we use the binomial series expansion for negative exponents:

\((1 + u)^{n} = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots\)

Here, \(u = 2x\) and \(n = -3\).

First term: \(1\).

Second term: \(-3(2x) = -6x\).

Third term: \(\frac{-3(-4)}{2}(2x)^2 = \frac{12}{2}(4x^2) = 24x^2\).

Thus, the expansion up to \(x^2\) is \(1 - 6x + 24x^2\).

(i) The first two terms of the expansion of \((1 + ax)^{\frac{2}{3}}\) are \(1 + \frac{2}{3}ax\). The expression for the coefficient of \(x\) in the expansion of \((1 + 2x)(1 + ax)^{\frac{2}{3}}\) is:

\(2 \cdot 1 + 1 \cdot \frac{2}{3}a = 0\)

Solving for \(a\):

\(2 + \frac{2}{3}a = 0\)

\(\frac{2}{3}a = -2\)

\(a = -3\)

(ii) Substitute \(a = -3\) into \((1 + ax)^{\frac{2}{3}}\) to get \((1 - 3x)^{\frac{2}{3}}\). The unsimplified terms in \(x^2\) and \(x^3\) are obtained from the expansion:

\((1 - 3x)^{\frac{2}{3}} = 1 - 2x + \frac{10}{3}x^2 - \frac{20}{9}x^3 + \ldots\)

Multiply by \(1 + 2x\) to find the term in \(x^3\):

\((1 + 2x)(1 - 2x + \frac{10}{3}x^2 - \frac{20}{9}x^3)\)

The term in \(x^3\) is:

\(-2 \cdot \frac{10}{3} + 1 \cdot \left(-\frac{20}{9}\right) = -\frac{20}{3} - \frac{20}{9}\)

Combine the fractions:

\(-\frac{60}{9} - \frac{20}{9} = -\frac{80}{9}\)

Thus, the term in \(x^3\) is \(-\frac{10}{3}x^3\).

To expand \((1 + x) \sqrt{(1 - 2x)}\), we first expand \(\sqrt{(1 - 2x)}\) using the binomial series for \((1 + u)^n\) where \(n = \frac{1}{2}\) and \(u = -2x\).

The expansion is:

\(1 + \frac{1}{2}(-2x) + \frac{1}{2} \left( \frac{1}{2} - 1 \right) \frac{(-2x)^2}{2!} + \ldots\)

Simplifying, we get:

\(1 - x + \frac{1}{2}(-2x)^2 \cdot \frac{-1}{2} = 1 - x + \frac{1}{2} \cdot 2x^2 = 1 - x + x^2\)

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

\((1 + x)(1 - x + x^2) = 1 - x + x^2 + x - x^2 + x^3\)

Combine like terms:

\(1 + 0x - \frac{3}{2}x^2 + x^3\)

Thus, the expansion up to \(x^2\) is:

\(1 - \frac{3}{2}x^2\)