The following function are continuous on $\left(0,\pi \right):$

The  function $f\left(x\right)$ $=\tan x$  is not defined at $x=\frac{\pi }{2},$ so $f$  is not continuous on $\left(0,\pi \right)$  since, the function $g\left(x\right)=x\sin \frac{1}{x}$  is continuous on $\left(0,\pi \right)$  and  the integral function of a continuous function is continuous, therefore $F\left(x\right)={\displaystyle {\int }_0^{x}t\sin \frac{1}{t}dt}$  is continuous on $\left(0,\pi \right)$ .

For the function  $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}1, 0<x\le \frac{3\pi }{4}\\ 2\sin \frac{2}{9}x,\frac{3\pi }{4}<x<\pi \end{array}\\ \end{array}\right.$

$\underset{x\to \frac{3\pi }{4}-}{\lim }f\left(x\right)=1,$  and $\underset{x\to \frac{3\pi }{4}+}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\frac{3\pi }{4}+h\right)$

                                              $=\underset{h\to 0}{\lim }2\sin \frac{2}{9}\left(\frac{3\pi }{4}+h\right)$

                                              $=2\sin \pi /6=1$  and $f\left(3\pi /4\right)=1$

$\therefore f\left(x\right)$  is continuous at $x=3\pi /4$  evidently

$f\left(x\right)$  is continuous at all other points.

For the function

$F\left(x\right)=\left\{\begin{array}{l}\begin{array}{l}x\sin x,0<x\le \frac{\pi }{2}\\ \frac{\pi }{2}\sin (\pi +x),\frac{\pi }{2}<x<\pi \end{array}\\ \end{array}\right.$

$\therefore F\left(\pi /2\right)=\frac{\pi }{2}\sin \frac{\pi }{2}=\frac{\pi }{2}$

$\underset{x\to \frac{\pi }{2}-}{\lim }F\left(x\right)=\underset{h\to 0}{\lim }F\left(\frac{\pi }{2}-h\right)$

                  $=\underset{h\to 0}{\lim }\left(\frac{\pi }{2}-h\right)\sin \left(\frac{\pi }{2}-h\right)$

                     $=\frac{\pi }{2}\sin \frac{\pi }{2}$

                      $=\frac{\pi }{2}$

$\underset{x\to \frac{\pi }{2}+}{\lim }F\left(x\right)=\underset{h\to 0}{\lim }F\left(\frac{\pi }{2}+h\right)$

                     $=\underset{h\to 0}{\lim }\frac{\pi }{2}\sin \left(\frac{3\pi }{2}+h\right)$

                   $=\frac{\pi }{2}\sin \frac{3\pi }{2}=-\frac{\pi }{2}$

$\therefore F\left(x\right)$  is not continuous at $x=\frac{\pi }{2}$