(i) Differentiate the curves: \(y = 4 \cos \frac{1}{2} x\) gives \(\frac{dy}{dx} = -2 \sin \frac{1}{2} x\) and \(y = \frac{1}{4-x}\) gives \(\frac{dy}{dx} = \frac{1}{(4-x)^2}\).

For perpendicular tangents, the product of the derivatives is \(-1\):

\(-2 \sin \frac{1}{2} a \cdot \frac{1}{(4-a)^2} = -1\)

Rearrange to find \(a\):

\(2 \sin \frac{1}{2} a = (4-a)^2\)

\(a = 4 - \sqrt{2 \sin \frac{1}{2} a}\)

(ii) Calculate values at \(a = 2\) and \(a = 3\):

For \(a = 2\), \(2 < 2.7027 \ldots\)

For \(a = 3\), \(3 > 2.587 \ldots\)

Thus, \(a\) lies between 2 and 3.

(iii) Use the iterative formula \(a_{n+1} = 4 - \sqrt{2 \sin \frac{1}{2} a_n}\):

Start with \(a_1 = 2\):

\(a_2 = 2.70272\)

\(a_3 = 2.60285\)

\(a_4 = 2.61152\)

\(a_5 = 2.61070\)

\(a_6 = 2.61077\)

Final answer: \(a = 2.611\)