(a) We start by rewriting \((2 - x^2)^{-2}\) as \(\left(\frac{1}{2}(1 - \frac{x^2}{2})\right)^{-2}\).

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

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

Substitute \(n = -2\) and \(u = -\frac{x^2}{2}\):

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

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

Multiply by \(\frac{1}{4}\) to account for the factor \(\frac{1}{4}\) from \(\left(\frac{1}{2}\right)^{-2}\):

\(\frac{1}{4}(1 + x^2 + \frac{3}{4}x^4) = \frac{1}{4} + \frac{1}{4}x^2 + \frac{3}{16}x^4\)

(b) The expansion is valid for \(|x| < \sqrt{2}\) because the binomial expansion is valid for \(|u| < 1\), where \(u = \frac{x^2}{2}\), leading to \(|x^2/2| < 1\), or \(|x| < \sqrt{2}\).