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

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

Here, \(u = \frac{x^2}{2}\) and \(n = 2\).

First term: \(\frac{1}{4}\).

Second term: \(-2 \times \frac{x^2}{2} = -x^2\).

Third term: \(\frac{2 \times 3}{2} \left(\frac{x^2}{2}\right)^2 = \frac{3}{16}x^4\).

Thus, the expansion up to \(x^4\) is \(\frac{1}{4} - \frac{1}{4}x^2 + \frac{3}{16}x^4\).

To expand \((1 - 3x)^{-\frac{1}{3}}\), we use the binomial series expansion for negative fractional powers:

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

Here, \(u = -3x\) and \(n = -\frac{1}{3}\).

First term: \(1\)

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

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

Fourth term: \(\frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6}(-3x)^3 = \frac{14}{3}x^3\)

Thus, the expansion up to \(x^3\) is \(1 + x + 2x^2 + \frac{14}{3}x^3\).

The binomial expansion of \(\sqrt{1 + 4x}\) is given by:

\(1 + 2x - 2x^2 + \cdots\)

Now, consider the expression \((a + bx)(1 + 2x - 2x^2)\).

Expanding this, we have:

\((a + bx)(1 + 2x - 2x^2) = a(1 + 2x - 2x^2) + bx(1 + 2x - 2x^2)\)

\(= a + 2ax - 2ax^2 + bx + 2bx^2 - 2bx^3\)

Collecting terms, the coefficient of \(x\) is \(2a + b\), and the coefficient of \(x^2\) is \(-2a + 2b\).

We are given:

\(2a + b = 3\)

\(-2a + 2b = -6\)

Solving these equations simultaneously:

From \(2a + b = 3\), we have \(b = 3 - 2a\).

Substitute \(b = 3 - 2a\) into \(-2a + 2b = -6\):

\(-2a + 2(3 - 2a) = -6\)

\(-2a + 6 - 4a = -6\)

\(-6a + 6 = -6\)

\(-6a = -12\)

\(a = 2\)

Substitute \(a = 2\) back into \(b = 3 - 2a\):

\(b = 3 - 2(2) = 3 - 4 = -1\)

Thus, \(a = 2\) and \(b = -1\).

To expand \((1 + 3x)^{\frac{2}{3}}\), we use the binomial series expansion for fractional powers:

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

Here, \(u = 3x\) and \(n = \frac{2}{3}\).

First term: \(1\).

Second term: \(\frac{2}{3} \cdot 3x = 2x\).

Third term: \(\frac{\frac{2}{3}(\frac{2}{3}-1)}{2} \cdot (3x)^2 = -x^2\).

Fourth term: \(\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{6} \cdot (3x)^3 = \frac{4}{3}x^3\).

Thus, the expansion up to \(x^3\) is \(1 + 2x - x^2 + \frac{4}{3}x^3\).

(a) To expand \(\sqrt[3]{1 + 6x}\), we use the binomial series expansion for \((1 + u)^n\) where \(n = \frac{1}{3}\) and \(u = 6x\).

The expansion is given by:

\(1 + \frac{n}{1!}u + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots\)

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

\(1 + \frac{1}{3}(6x) + \frac{1}{3} \cdot \frac{-2}{3} \cdot \frac{(6x)^2}{2} + \frac{1}{3} \cdot \frac{-2}{3} \cdot \frac{-5}{3} \cdot \frac{(6x)^3}{6}\)

Simplifying each term:

First term: \(1\)

Second term: \(2x\)

Third term: \(-4x^2\)

Fourth term: \(\frac{40}{3}x^3\)

Thus, the expansion up to \(x^3\) is \(1 + 2x - 4x^2 + \frac{40}{3}x^3\).

(b) The expansion is valid for \(|6x| < 1\), which simplifies to \(|x| < \frac{1}{6}\).