(i) We start by simplifying \((\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x})\).

This is a difference of squares: \((a+b)(a-b) = a^2 - b^2\).

Thus, \((\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 2x\).

Therefore, \((\sqrt{1+x} + \sqrt{1-x})(\sqrt{1+x} - \sqrt{1-x}) = 2x\).

Now, deduce that:

\(\frac{1}{\sqrt{1+x} + \sqrt{1-x}} = \frac{\sqrt{1+x} - \sqrt{1-x}}{2x}.\)

(ii) To find the expansion of \(\frac{1}{\sqrt{1+x} + \sqrt{1-x}}\), we use the binomial expansion for \(\sqrt{1+x}\) and \(\sqrt{1-x}\).

\(\sqrt{1+x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}\)

\(\sqrt{1-x} \approx 1 - \frac{x}{2} - \frac{x^2}{8}\)

Adding these gives:

\(\sqrt{1+x} + \sqrt{1-x} \approx 2 - \frac{x^2}{4}\)

Thus, \(\frac{1}{\sqrt{1+x} + \sqrt{1-x}} \approx \frac{1}{2 - \frac{x^2}{4}}\).

Using the binomial expansion for \(\frac{1}{1-y}\), where \(y = \frac{x^2}{8}\), we get:

\(\frac{1}{2} \left(1 + \frac{x^2}{8}\right)^{-1} \approx \frac{1}{2} \left(1 - \frac{x^2}{8}\right)\)

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