(a) Sketch \(y=\ln x\) and \(y=\operatorname{cosec}(x/2)\) on \(0<x<\pi\):

On this interval, \(\ln x\) is strictly increasing. Since \(\sin(x/2)\) is strictly increasing and positive, \(\operatorname{cosec}(x/2)\) is strictly decreasing. Thus there can be at most one intersection.

Near zero, \(\ln x\to-\infty\) and \(\operatorname{cosec}(x/2)\to+\infty\). At \(x=\pi\), \(\ln\pi>1=\operatorname{cosec}(\pi/2)\). Continuity guarantees an intersection, so there is exactly one root.

(b) Work in radians and set \(f(x)=\ln x-1/\sin(x/2)\).

\[f(2.6)=-0.082308\ldots<0,\qquad f(2.9)=0.057370\ldots>0.\]

Since the function is continuous and changes sign, the root lies between 2.6 and 2.9.

(c) Start with \(x_0=2.6\) and use the given iteration in radians. Keep full precision internally and display each result to 5 decimal places.

The iterates agree to the required precision, so the root is \(\boxed{x=2.767}\) to 3 decimal places.