a) Function: \( f(x) = 2 – \frac{12}{x + 5} \), domain: \( -7 \leq x \leq 7 \), \( x \neq -5 \).
At \( x = -7 \): \( f(-7) = 2 – \frac{12}{-2} = 2 + 6 = 8 \).
At \( x = 7 \): \( f(7) = 2 – \frac{12}{12} = 2 – 1 = 1 \).
As \( x \to -5^- \), \( x + 5 \to 0^- \), \( \frac{12}{x + 5} \to -\infty \), so \( f(x) \to +\infty \).
As \( x \to -5^+ \), \( x + 5 \to 0^+ \), \( \frac{12}{x + 5} \to +\infty \), so \( f(x) \to -\infty \).
Derivative: \( f'(x) = \frac{12}{(x + 5)^2} > 0 \), so \( f \) is increasing.
On \([-7, -5)\): \( f(x) \) from 8 to \( +\infty \).
On \((-5, 7]\): \( f(x) \) from \( -\infty \) to 1.
Range: \( (-\infty, 1] \cup [8, +\infty) \) or \( f(x) \leq 1 \) or \( f(x) \geq 8 \). [3]

b) Set \( y = 2 – \frac{12}{x + 5} \).
Swap: \( x = 2 – \frac{12}{y + 5} \).
Solve: \( x – 2 = -\frac{12}{y + 5} \implies 2 – x = \frac{12}{y + 5} \implies y + 5 = \frac{12}{2 – x} \implies y = \frac{12}{2 – x} – 5 \).
Inverse: \( f^{-1}(x) = \frac{12}{2 – x} – 5 \). [3]

c) Range of \( f^{-1}(x) \): Domain of \( f \), i.e., \( -7 \leq f^{-1}(x) \leq 7 \), \( f^{-1}(x) \neq -5 \). [1]