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:

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