(b) Start with \(y = \frac{-x}{\sqrt{4-x^2}}\). Square both sides to get \(x^2 = y^2(4-x^2)\).

Rearrange to \(x^2(1+y^2) = 4y^2\).

Solving for \(x\), we get \(x = \frac{\pm 2y}{\sqrt{1+y^2}}\).

Select the correct square root: \(f^{-1}(x) = \frac{-2x}{\sqrt{1+x^2}}\).

(c) The maximum possible value of \(a\) is 1, as \(-1 < x < 1\) is required for \(fg\) to be formed.

(d) Substitute \(g(x) = 2x\) into \(f(x)\): \(fg(x) = f(2x) = \frac{-2x}{\sqrt{4-(2x)^2}}\).

Simplify to \(fg(x) = \frac{-x}{\sqrt{1-x^2}}\).