$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right): \frac{1-\sin x}{(\pi-2 x)^3} \\ & \text { On putting } x=\frac{\pi}{2}-y, y \rightarrow 0 \\ & =\lim _{y \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\frac{\pi}{2}-y}{2}\right) \cdot\left(1-\sin \left(\frac{\pi}{2}-y\right)\right.}{\left(\pi-2\left(\frac{\pi}{2}-y\right)\right)^3} \\ & =\lim _{y \rightarrow 0} \frac{\left(\tan \frac{y}{2}\right) \cdot(1-\cos y)}{8 y^3} \\ & =\lim _{y \rightarrow 0} \frac{\left(\tan \frac{y}{2}\right) \cdot 2 \cdot\left(\sin \frac{y}{2}\right)\left(\sin \frac{y}{2}\right)}{64 \cdot\left(\frac{y}{2}\right)\left(\frac{y}{2}\right)\left(\frac{y}{2}\right)} \\ & =\frac{2}{64}=\frac{1}{32} \\ & \end{aligned}$