(a) The vertical asymptote occurs where the denominator is zero: \(x + a = 0 \Rightarrow x = -a\). For the oblique asymptote, divide \(x^2 + a\) by \(x + a\) to get \(y = x - a\).

(b) Differentiate \(y = \frac{x^2 + a}{x + a}\) to find stationary points: \(\frac{dy}{dx} = \frac{(x+a)^2 - (x^2+a)}{(x+a)^2} = \frac{-a^2}{(x+a)^2}\). Set \(\frac{dy}{dx} = 0\) to find \(x = -a \pm \sqrt{a^2 + a}\).

(c) Sketch the curve showing the asymptotes \(x = -a\) and \(y = x - a\). The curve intersects the y-axis at \((0, 1)\).

(d) Sketch the absolute value of the curve from (c), reflecting any negative parts above the x-axis.

(e) Solve \(\left| \frac{x^2 + a}{x + a} \right| = a\). This leads to \(x^2 - ax + a - a^2 = 0\). The discriminant \(a^2 - 4(a-a^2) = 5a^2 - 4a > 0\) gives \(a > \frac{3}{4}\).