(a) Sketch the graphs

\[ y=\ln x \]

and

\[ y=\operatorname{cosec}\frac12x \]

for \(0<x<\pi\).

On this interval, \(y=\ln x\) is increasing.

Also, \(\sin\frac12x\) is increasing on \(0<x<\pi\), so \(y=\operatorname{cosec}\frac12x\) is decreasing.

Therefore the two graphs can meet at most once.

Since the sketch shows one intersection, the equation has exactly one root in \(0<x<\pi\).

(b) Let

\[ f(x)=\ln x-\operatorname{cosec}\frac12x. \]

At \(x=2.6\),

\[ f(2.6)=\ln(2.6)-\operatorname{cosec}(1.3)=-0.0823\ldots \]

At \(x=2.9\),

\[ f(2.9)=\ln(2.9)-\operatorname{cosec}(1.45)=0.0574\ldots \]

Since there is a change of sign, the root lies between \(2.6\) and \(2.9\).

(c) Using

\[ x_{n+1}=\exp\left(\operatorname{cosec}\frac12x_n\right), \]

and starting with \(x_0=2.6\):

\[ x_1=2.82306 \]

\[ x_2=2.75335 \]

\[ x_3=2.77082 \]

\[ x_4=2.76609 \]

\[ x_5=2.76734 \]

\[ x_6=2.76701 \]

\[ x_7=2.76710 \]

Therefore the root correct to \(3\) decimal places is

\[ x=2.767. \]

Answer: \(2.767\).