We start by using the binomial expansion for \((1 + u)^n\), where \(u = -2x\) and \(n = -\frac{1}{2}\).

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 \(u = -2x\) and \(n = -\frac{1}{2}\), we have:

\((1 - 2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{-\frac{1}{2}(-\frac{3}{2})}{2!}(-2x)^2 + \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{3!}(-2x)^3 + \cdots\)

Calculating each term:

First term: \(1\)

Second term: \(\frac{1}{2} \times 2x = x\)

Third term: \(\frac{3}{8} \times 4x^2 = \frac{3}{2}x^2\)

Fourth term: \(\frac{15}{48} \times 8x^3 = \frac{5}{2}x^3\)

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

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

To find the coefficients \(a\) and \(b\), we expand \(\sqrt[3]{(1 + 9x)}\) using the binomial series for small \(x\):

\(\sqrt[3]{(1 + 9x)} = (1 + 9x)^{1/3} \approx 1 + \frac{1}{3}(9x) - \frac{1}{9}(9x)^2 + \frac{5}{81}(9x)^3\)

\(= 1 + 3x - 9x^2 + 45x^3\)

Comparing this with \(1 + 3x + ax^2 + bx^3\), we find:

\(a = -9\)

\(b = 45\)

To solve this, we need to expand both expressions for small values of \(x^2\) and find the coefficient of \(x^4\).

First, consider \((1 - 2x^2)^{-2}\). Using the binomial expansion for \((1 - u)^{-n}\), we have:

\((1 - 2x^2)^{-2} = 1 + 4x^2 + 12x^4 + \cdots\)

Next, consider \((1 + 6x^2)^{\frac{2}{3}}\). Using the binomial expansion for \((1 + u)^{n}\), we have:

\((1 + 6x^2)^{\frac{2}{3}} = 1 + 4x^2 - 4x^4 + \cdots\)

Subtracting these expansions:

\((1 - 2x^2)^{-2} - (1 + 6x^2)^{\frac{2}{3}} = (1 + 4x^2 + 12x^4) - (1 + 4x^2 - 4x^4)\)

\(= 12x^4 + 4x^4 = 16x^4\)

Thus, \(k = 16\).

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

The binomial expansion is given by:

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

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

\(1 + \left(-\frac{1}{3}\right)(3x) + \frac{-\frac{1}{3}(-\frac{4}{3})}{2}(3x)^2 + \frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6}(3x)^3\)

Calculating each term:

First term: \(1\)

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

Third term: \(\frac{-\frac{1}{3} \times -\frac{4}{3}}{2} \times 9x^2 = 2x^2\)

Fourth term: \(\frac{-\frac{1}{3} \times -\frac{4}{3} \times -\frac{7}{3}}{6} \times 27x^3 = -\frac{14}{3}x^3\)

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

\(1 - x + 2x^2 - \frac{14}{3}x^3\)

To expand \(\frac{1 + 3x}{\sqrt{1 + 2x}}\), we first consider the expansion of \((1 + 2x)^{-\frac{1}{2}}\).

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

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

Now, multiply this expansion by \(1 + 3x\):

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

Combine like terms:

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

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