(i) To differentiate \(y = \csc x = \frac{1}{\sin x}\), use the quotient rule or chain rule. The derivative of \(\sin x\) is \(\cos x\), so:

\(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{\sin x} \right) = -\frac{\cos x}{\sin^2 x} = -\csc x \cot x.\)

(ii) The gradients of the curves are equal when \(x = a\). For \(y = \csc x\), the gradient is \(-\csc a \cot a\). For \(y = e^{-x}\), the gradient is \(-e^{-a}\). Equating these gives:

\(-e^{-a} = -\csc a \cot a.\)

Rearranging gives:

\(a = \arctan \left( \frac{e^a}{\sin a} \right).\)

(iii) To verify \(a\) lies between 1 and 1.5, calculate the expression for \(a = 1\) and \(a = 1.5\). Check that the values satisfy the inequality.

(iv) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{e^{a_n}}{\sin a_n} \right)\). Start with an initial guess, such as \(a_0 = 1\), and iterate:

\(a_1 = \arctan \left( \frac{e^{1}}{\sin 1} \right) \approx 1.317\)

Continue iterating until the value stabilizes to 3 decimal places:

\(a_2 \approx 1.317\)

\(a_3 \approx 1.317\)

The final answer is \(a \approx 1.317\).