(a) The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\). The leading coefficients of the numerator and denominator are equal, so \(y = 1\) is the horizontal asymptote.

(b) Consider \(y = \frac{x^2 + x - 4}{x^2 + x + 2}\). Rearrange to form a quadratic in \(x\): \((y-1)x^2 + (y-1)x + (2y + 4) = 0\). The discriminant \(\Delta = (y-1)^2 - 4(y-1)(2y+4) \geq 0\) gives \((y-1)(7y + 17) \leq 0\). Solving this inequality gives \(-\frac{17}{7} \leq y < 1\).

(c) Find stationary points by setting \(\frac{dy}{dx} = 0\). Differentiate: \(\frac{dy}{dx} = \frac{12x + 6}{(x^2 + x + 2)^2} = 0\). Solving gives \(x = -\frac{1}{2}\). Substitute back to find \(y = -\frac{17}{7}\). Stationary point is \(\left(-\frac{1}{2}, -\frac{17}{7}\right)\).

(d) Sketch the curve, noting the asymptote \(y = 1\) and intersections at \((0, -2), \left(\frac{1}{2}(1-\sqrt{7}), 0\right), \left(\frac{1}{2}(1+\sqrt{7}), 0\right)\).

(e) For \(y = \frac{|x|^2 + |x| - 4}{|x|^2 + |x| + 2} < -\frac{1}{2}\), solve \(3|x|^2 + 3|x| - 6 = 0\). Critical points are \(|x| = 1\). Thus, \(-1 < x < 1\).