First, consider the expansion of \(\sqrt{1 + 4x} = (1 + 4x)^{1/2}\).

Using the binomial expansion formula, the general term is given by:

\(\binom{1/2}{n} (4x)^n = \frac{1}{2} \cdot \frac{1}{2} - 1 \cdot \frac{1}{2} - 2 \cdots \frac{1}{2} - (n-1) \cdot \frac{(4x)^n}{n!}\)

For \(x^3\), we need the term where \(n = 3\):

\(\frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{(4x)^3}{6} = \frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{64x^3}{6} = 4x^3\)

Next, consider the term in \(x^2\) for \((1 + 4x)^{1/2}\):

\(\frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{(4x)^2}{2} = \frac{1}{2} \cdot \frac{-1}{2} \cdot 16x^2 = -2x^2\)

Now, multiply \((3 + x)\) by these terms:

\((3 + x)(4x^3) = 12x^3 + 4x^4\)

\((3 + x)(-2x^2) = -6x^2 - 2x^3\)

Combine the \(x^3\) terms:

\(12x^3 - 2x^3 = 10x^3\)

Thus, the coefficient of \(x^3\) is 10.

To expand \((1 - 4x)^{\frac{1}{4}}\), we use the binomial series expansion for \((1 + u)^n\), which is:

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

Here, \(n = \frac{1}{4}\) and \(u = -4x\).

The first term is \(1\).

The second term is \(\frac{1}{4}(-4x) = -x\).

The third term is \(\frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{(-4x)^2}{2} = -\frac{3}{2}x^2\).

The fourth term is \(\frac{1}{4} \cdot \frac{-3}{4} \cdot \frac{-7}{4} \cdot \frac{(-4x)^3}{6} = -\frac{7}{2}x^3\).

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

\(1 - x - \frac{3}{2}x^2 - \frac{7}{2}x^3\).

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

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

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

First term: \(\frac{1}{27}\)

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

Third term: \(\frac{3 \times 4}{2} \left(\frac{2}{3}x\right)^2 = \frac{8}{9}x^2\)

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

\(\frac{1}{27} - \frac{2}{27}x + \frac{8}{81}x^2\)

We need to expand \((1 + 6x)^{-\frac{1}{3}}\) using the binomial series expansion for negative exponents.

The binomial series for \((1 + u)^n\) is given by:

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

Here, \(n = -\frac{1}{3}\) and \(u = 6x\).

First term: \(1\)

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

Third term: \(\frac{-\frac{1}{3}(-\frac{4}{3})}{2} \times (6x)^2 = 8x^2\)

Fourth term: \(\frac{-\frac{1}{3}(-\frac{4}{3})(-\frac{7}{3})}{6} \times (6x)^3 = -\frac{112}{3}x^3\)

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

\(1 - 2x + 8x^2 - \frac{112}{3}x^3\)

To expand \((1+2x)^{-\frac{3}{2}}\), 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 \(n = -\frac{3}{2}\).

The first two terms are:

\(1 + \left(-\frac{3}{2}\right)(2x) = 1 - 3x\).

The term in \(x^2\) is:

\(\frac{-\frac{3}{2}(-\frac{3}{2}-1)}{2!}(2x)^2 = \frac{-\frac{3}{2}(-\frac{5}{2})}{2}(4x^2) = \frac{15}{4}x^2\).

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

\(1 - 3x + \frac{15}{4}x^2\).

Now, multiply by \((2-x)\):

\((2-x)(1 - 3x + \frac{15}{4}x^2) = 2 - 3x + \frac{15}{4}x^2 - x + 3x^2 - \frac{15}{4}x^3\).

Combine like terms up to \(x^2\):

\(2 - 4x + \left(\frac{15}{4} + 3\right)x^2 = 2 - 4x + \frac{27}{4}x^2\).

Simplify the coefficients:

\(2 - 7x + 18x^2\).