$\underset{x\to \frac{\pi }{2}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{\pi }{4}-\frac{x}{2}\right)\left(1-\mathrm{s}\mathrm{i}\mathrm{n}x\right)}{{\left(\pi -2x\right)}^3}$

Let, $x=\frac{\pi }{2}+y$

 $\Rightarrow \underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}=\frac{\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{-y}{2}\right)\left(1-\mathrm{c}\mathrm{o}\mathrm{s}y\right)}{{\left(-2y\right)}^3}$

$$\begin{gathered} =\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{-\mathrm{t}\mathrm{a}\mathrm{n}\frac{y}{2}\cdot 2\mathrm{s}\mathrm{i}{\mathrm{n}}^2\frac{y}{2}}{\left(-8\right)y^3}=\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{32}\frac{\mathrm{t}\mathrm{a}\mathrm{n}\frac{y}{2}}{\left(\frac{y}{2}\right)}\cdot {\left[\frac{\mathrm{s}\mathrm{i}\mathrm{n}\frac{y}{2}}{\frac{y}{2}}\right]}^2 \\ =\underset{y\to 0}{\lim }\frac{1}{32}\times \frac{\tan \frac{y}{2}}{\left(\frac{y}{2}\right)}.{\left(\underset{y\to 0}{\lim }\frac{\sin \frac{y}{2}}{\frac{y}{2}}\right)}^2 \end{gathered}$$

$\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=\underset{x\to 0}{\lim }\frac{\tan x}{x}=1$

$\Rightarrow \frac{1}{32}\times 1\times 1^2$

$=\frac{1}{32}$