$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}{\left(\frac{3}{2}\right)}^{\frac{\cot 3x}{\cot 2x}}; 0<x<\frac{\pi }{2}\\ b+3,;x=\frac{\pi }{2} iscontinuousat x=\frac{\mathrm{\pi }}{2}\\ {(1+\left|\cot x\right|)}^{\left(a\left|\tan x\right|\right)/b};\frac{\pi }{2}<x<\pi \end{array}, then\\ \end{array}\right.$

$f\left(\frac{{\pi }^{-}}{2}\right)=\underset{h\to 0}{\lim }{\left(\frac{3}{2}\right)}^{\frac{\cot \left(3(\frac{\pi }{2}-h)\right)}{\cot \left(2(\frac{\pi }{2}-h)\right)}}$ 

$=\underset{h\to 0}{\lim }{\left(\frac{3}{2}\right)}^{\frac{\tan 3h}{-\cot 2h}}=\underset{h\to 0}{\lim }{\left(\frac{3}{2}\right)}^{-\left(\tan 3h\right)\left(\tan 2h\right)}=1$

$f\left(\frac{{\pi }^{+}}{2}\right)=\underset{h\to 0}{\lim }{(1+\left|\cot (\frac{\pi }{2}+h)\right|)}^{\frac{\left(a\left|\tan \right(\frac{\pi }{2}+h\left)\right|\right)}{b}}$

$=\underset{h\to 0}{\lim }{\left(1+\tanh \right)}^{\frac{a\coth }{b}}=e^{\underset{h\to 0}{\lim }\left(1+\tanh -1\right)\frac{a\coth }{b}}=e^{a/b}$

$f\left(\frac{\pi }{2}\right)=b+3$

$f\left(x\right)$ is continuous at$x=\pi /2\Rightarrow 1=b+3=e^{a/b}\Rightarrow b=-2$  and $a=0$