(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + 1 = 0\). This has no real roots, so there are no vertical asymptotes. The horizontal asymptote is found by considering the limits as \(x \to \pm \infty\), giving \(y = 1\).

(b) Rewrite \(y = 1 + \frac{2}{x^2 + 1}\). Since \(\frac{2}{x^2 + 1} > 0\), it follows that \(y > 1\). The maximum value of \(y\) occurs when \(\frac{2}{x^2 + 1}\) is maximized, which is 2, giving \(y = 3\). Thus, \(1 < y \leq 3\).

(c) Differentiate \(y = \frac{x^2 + 3}{x^2 + 1}\) to find stationary points: \(\frac{d}{dx} \left( \frac{x^2 + 3}{x^2 + 1} \right) = \frac{(x^2 + 1)(2x) - (x^2 + 3)(2x)}{(x^2 + 1)^2} = 0\). Solving gives \(x = 0\), and substituting back gives \(y = 3\). Thus, the stationary point is \((0, 3)\).

(d) The sketch shows the curve with a horizontal asymptote at \(y = 1\) and a stationary point at \((0, 3)\). The curve intersects the y-axis at \((0, 3)\).

(e) For \(y = \frac{x^2 + 1}{x^2 + 3} < \frac{1}{2}\), solve \(2(x^2 + 1) < x^2 + 3\), giving \(x^2 < 1\). Thus, \(-1 < x < 1\).