(a) To find the vertical asymptotes, set the denominator equal to zero: \(2x^2 - 7x + 3 = 0\). Solving gives \(x = \frac{1}{2}\) and \(x = 3\). The horizontal asymptote is found by considering the degrees of the polynomials: \(y = \frac{4}{2} = 2\).

(b) To find stationary points, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{(2x^2 - 7x + 3)(8x + 1) - (4x^2 + x + 1)(4x - 7)}{(2x^2 - 7x + 3)^2}\).

Set \(\frac{dy}{dx} = 0\) and solve \(-3x^2 + 2x + 1 = 0\) to find \(x = -\frac{1}{3}\) and \(x = 1\). Substitute back to find \(y\) values: \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) and \((1, -3)\).

(c) Sketch the curve using the asymptotes and stationary points. The curve intersects the y-axis at \((0, \frac{1}{3})\).

(d) For \(\left| \frac{4x^2 + x + 1}{2x^2 - 7x + 3} \right| = k\) to have 4 distinct real solutions, \(k > 3\).