(a) To find \(f(x)\), integrate \(f'(x) = 2x^2 - 7 - \frac{4}{x^2}\):

\(f(x) = \int (2x^2 - 7 - \frac{4}{x^2}) \, dx = \frac{2}{3}x^3 - 7x + 4x^{-1} + c\).

Given \(f(1) = -\frac{1}{3}\), substitute to find \(c\):

\(-\frac{1}{3} = \frac{2}{3}(1)^3 - 7(1) + 4(1)^{-1} + c\)

\(-\frac{1}{3} = \frac{2}{3} - 7 + 4 + c\)

\(-\frac{1}{3} = -\frac{1}{3} + c\)

\(c = 2\)

Thus, \(f(x) = \frac{2}{3}x^3 - 7x + 4x^{-1} + 2\).

(b) Stationary points occur where \(f'(x) = 0\):

\(2x^2 - 7 - \frac{4}{x^2} = 0\)

\(2x^4 - 7x^2 - 4 = 0\)

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

\(x^2 = 4 \Rightarrow x = \pm 2\)

For \(x = 2\), \(f(2) = \frac{2}{3}(2)^3 - 7(2) + 4(2)^{-1} + 2 = \frac{14}{3}\)

For \(x = -2\), \(f(-2) = \frac{2}{3}(-2)^3 - 7(-2) + 4(-2)^{-1} + 2 = -\frac{26}{3}\)

Coordinates: \((2, \frac{14}{3})\) and \((-2, -\frac{26}{3})\).

(c) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = \frac{d}{dx}(2x^2 - 7 - \frac{4}{x^2}) = 4x + 8x^{-3}\).

(d) Evaluate \(f''(x)\) at the stationary points:

\(f''(2) = 4(2) + 8(2)^{-3} = 9 > 0\), so minimum at \(x = 2\).

\(f''(-2) = 4(-2) + 8(-2)^{-3} = -9 < 0\), so maximum at \(x = -2\).