(a) The vertical asymptote occurs where the denominator is zero: \(x + 2 = 0 \Rightarrow x = -2\).

The horizontal asymptote is found by dividing the leading coefficients: \(y = \frac{x^2 + ax + 1}{x + 2} = x + a - 2\).

(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\).

Set \(\frac{dy}{dx} = 0\):

\(x^2 + 4x + 2a - 1 = 0\).

The discriminant \(16 - 4(2a-1) = 20 - 8a < 0\) implies no real roots, hence no stationary points.

(c) The curve intersects the \(y\)-axis at \(x = 0\):

\(y = \frac{0^2 + a(0) + 1}{0 + 2} = \frac{1}{2}\).

Asymptotes: vertical at \(x = -2\), oblique at \(y = x + a - 2\).

(d)(i) Sketch \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes.

(ii) Draw the line \(y = a\) on the sketch.

(iii) Solve \(\left| \frac{x^2 + ax + 1}{x + 2} \right| < a\) for given intervals:

\(\frac{x^2 + ax + 1}{x + 2} = a\) or \(\frac{x^2 + ax + 1}{x + 2} = -a\).

\(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\).

\(-a - \sqrt{a^2 - 2a - 1} < x < -\sqrt{2a - 1}\) and \(-a + \sqrt{a^2 - 2a - 1} < x < \sqrt{2a - 1}\).

Given intervals imply \(a = 5\).