(a) Calculate \(f(1) = \frac{e^{2} + 1}{e^{2} - 1}\) and \(f(1.5) = \frac{e^{3} + 1}{e^{3} - 1}\). Verify that \(f(1) < 1\) and \(f(1.5) > 1.5\), confirming \(a\) is between 1 and 1.5.

(b) Use the iterative formula \(x_{n+1} = \frac{e^{2x_n} + 1}{e^{2x_n} - 1}\). Start with \(x_1 = 1\):

\(x_2 = 1.2311\)

\(x_3 = 1.2032\)

\(x_4 = 1.2002\)

\(x_5 = 1.2000\)

Thus, \(a = 1.20\) to 2 decimal places.

(c) Use the quotient rule to find \(f'(x) = \frac{(e^{2x} - 1)(2e^{2x}) - (e^{2x} + 1)(2e^{2x})}{(e^{2x} - 1)^2}\).

Simplify to \(f'(x) = \frac{-4e^{2x}}{(e^{2x} - 1)^2}\).

Set \(f'(x) = -8\):

\(\frac{-4e^{2x}}{(e^{2x} - 1)^2} = -8\)

\(2(e^{2x})^2 - 5e^{2x} + 2 = 0\)

Solve the quadratic equation for \(e^{2x}\):

\(e^{2x} = 2\) or \(e^{2x} = \frac{1}{2}\)

Thus, \(x = \frac{1}{2} \ln 2\).