(a) The vertical asymptotes occur where the denominator is zero: \(4 - 4x^2 = 0\), giving \(x = 1\) and \(x = -1\). The horizontal asymptote is \(y = 0\) as \(x\) approaches infinity.

(b) To find stationary points, set \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{(4-4x^2)(4) - (4x+5)(-8x)}{(4-4x^2)^2}\). Simplify and solve \(16x^2 + 40x + 16 = 0\) to find \((-2, \frac{1}{4})\) and \(-\frac{1}{2}, 1\).

(c) The curve intersects the x-axis where \(y = 0\), giving \(x = -\frac{5}{4}\). It intersects the y-axis where \(x = 0\), giving \(y = \frac{5}{4}\).

(d) For \(y = \left| \frac{4x+5}{4-4x^2} \right|\), solve \(4|4x+5| > 5|4-4x^2|\). This gives two cases: \(\frac{4x+5}{4-4x^2} > \frac{5}{4}\) and \(\frac{4x+5}{4-4x^2} < -\frac{5}{4}\). Solving these inequalities gives the ranges \(\frac{2}{3} - \frac{1}{3}\sqrt{6} < x < -\frac{4}{3}\) and \(0 < x < \frac{2}{3} + \frac{1}{3}\sqrt{6}\), excluding \(x = \pm 1\).