Let $$f\left( x \right) = \frac{{x\left( {x - 2} \right)}}{{{x^2} - 4}} = \frac{{x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{x}{{x + 2}}$$
Since $$f\left( x \right)$$  is continuous at $$x = 2$$
$$\eqalign{ & \therefore \,\mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right) \cr & \Rightarrow \mathop {\lim }\limits_{x \to 2} \frac{x}{{x + 2}} = f\left( 2 \right) \cr & \Rightarrow f\left( 2 \right) = \frac{2}{4} = \frac{1}{2} \cr} $$

Clearly, $$f\left( x \right)$$  is not continuous at $$x = 0$$
Now, $$f\left( 1 \right) = \cos \,3$$
$$\eqalign{ & \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \,3\,.\cos \left\{ {\frac{\pi }{2}\left( { - h} \right)} \right\} = \cos \,3 \cr & \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \,3.\,\cos \left\{ {\frac{\pi }{2}\left( h \right)} \right\} = \cos \,3 \cr & {\text{Further }}f\left( 2 \right) = \cos \,1.\cos \frac{\pi }{2} = 0 \cr & \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \,1.\,\cos \left\{ {\frac{\pi }{2}\left( {1 - h} \right)} \right\} = \mathop {\lim }\limits_{h \to 0} \cos \,1.0 = 0 \cr & \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} \cos \,1.\,\cos \left\{ {\frac{\pi }{2}\left( {1 + h} \right)} \right\} = 0 \cr} $$
$$\therefore \,f\left( x \right)$$   is continuous at $$x = 1$$  and $$x = 2$$  both.

$$f\left( x \right) = \tan \,x$$   is discontinuous when $$x = \left( {2n + 1} \right)\frac{\pi }{2},\,n\, \in \,I$$
$$f\left( x \right) = x\left[ x \right]$$   is discontinuous when $$x = k,\,k\, \in \,I$$
$$f\left( x \right) = \sin \,\left[ {n\pi x} \right]$$    is discontinuous when $$n\pi x = k,\,k\, \in \,I$$
Thus, all the above functions have infinite number of points of discontinuity.
But $$f\left( x \right) = \frac{{\left| x \right|}}{x}$$   is discontinuous when $$x = 0$$  only.

We have, $$f\left( x \right) = \frac{1}{{1 - x}}$$
As at $$x = 1,\,f\left( x \right)$$   is not defined, $$x = 1$$  is a point of discontinuity of $$f\left( x \right).$$
If $$x \ne 1,\,f\left[ {f\left( x \right)} \right] = f\left( {\frac{1}{{1 - x}}} \right) = \frac{1}{{1 - \frac{1}{{\left( {1 - x} \right)}}}} = \frac{{x - 1}}{x}$$
$$\therefore \,x = 0,\,1$$   are points of discontinuity of $$f\left[ {f\left( x \right)} \right].$$
If $$x \ne 0,\,x \ne 1$$
$$f\left[ {f\left\{ {f\left( x \right)} \right\}} \right] = f\left( {\frac{{x - 1}}{x}} \right) = \frac{1}{{1 - \frac{{\left( {x - 1} \right)}}{x}}} = x$$

$$\eqalign{ & {\text{If }}n > 1,\,\sin \,x > {\sin ^n}x.{\text{ If }}0 < n < 1,\,\sin \,x < {\sin ^n}x\, \cr & \therefore {\text{ if }}n > 1,\,f\left( x \right) = \frac{{2\left( {\sin \,x - {{\sin }^n}x} \right) + \left| {\sin \,x - {{\sin }^n}x} \right|}}{{2\left( {\sin \,x - {{\sin }^n}x} \right) - \left| {\sin \,x - {{\sin }^n}x} \right|}} = 3 \cr & {\text{If }}0 < n < 1,\,f\left( x \right) = \frac{{2\left( {\sin \,x - {{\sin }^n}x} \right) - \left| {\sin \,x - {{\sin }^n}x} \right|}}{{2\left( {\sin \,x - {{\sin }^n}x} \right) + \left| {\sin \,x - {{\sin }^n}x} \right|}} = \frac{1}{3} \cr & \therefore {\text{ if }}n > 1,\,g\left( x \right) = 3,\,x\, \in \left( {0,\,\pi } \right)\, \cr} $$
$$\therefore \,g\left( x \right)$$   is continuous and differentiable at $$x = \frac{\pi }{2}$$
$$\eqalign{ & {\text{If }}\,0 < n < 1,\,g\left( x \right) = 0\,x\, \in \left( {0,\,\frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{2},\,\pi } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3,\,x = \frac{\pi }{2} \cr} $$
Then $$g\left( {\frac{\pi }{2} + 0} \right) = 0,\,\,g\left( {\frac{\pi }{2} - 0} \right) = 0,\,\,g\left( {\frac{\pi }{2}} \right) = 3$$
So, $$g\left( x \right)$$  is not continuous at $$x = \frac{\pi }{2}$$
Hence, $$g\left( x \right)$$  is also not differentiable at $$x = \frac{\pi }{2}.$$

$$\eqalign{ & \cos \,x + \sin \,x = \sqrt 2 \,\cos \,\left( {x - \frac{\pi }{4}} \right) \cr & {\text{So, }}f\left( x \right) = \left[ {\sqrt 2 \,\cos \left( {x - \frac{\pi }{4}} \right)} \right] \cr} $$
We know that $$\left[ x \right]$$ is discontinuous at integral values of $$x.$$
Now, $${\sqrt 2 \,\cos \left( {x - \frac{\pi }{4}} \right)}$$    is an integer at $$x = \frac{\pi }{2},\,\frac{\pi }{2} + \frac{\pi }{4},\,\pi + \frac{\pi }{4},\,\frac{{3\pi }}{2} + \frac{\pi }{4}$$

Let \[f\left( x \right) = \left\{ \begin{array}{l} \frac{{{x^3} - 3x + 2}}{{{{\left( {x - 1} \right)}^2}}},{\rm{ }}\forall {\rm{ }}x \ne 1\\ \,\,\,\,\,\,\,\,k,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ }}\forall {\rm{ }}x = 1 \end{array} \right.\]     and $$f\left( x \right)$$  is continuous.
$$\eqalign{ & \therefore \mathop {\lim }\limits_{x \to 1} f\left( x \right) = k \cr & \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 3x + 2}}{{{{\left( {x - 1} \right)}^2}}} = k \cr & \Rightarrow k = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 3}}{{2\left( {x - 1} \right)}}\,\,\,\,\,\,\,\,\,\left[ {{\text{By L'Hospital rule}}} \right] \cr & \Rightarrow k = \mathop {\lim }\limits_{x \to 1} \frac{{6x}}{2}\,\,\,\,\,\,\,\,\left[ {{\text{By L'Hospital rule}}} \right] \cr & \Rightarrow k = 3 \cr} $$

Given function is defined as : \[f\left( x \right) = \left\{ \begin{array}{l} {x^p}\cos \left( {\frac{1}{x}} \right),\,\,\,x \ne 0\\ \,\,\,\,\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \end{array} \right.\]
For continuity :
$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}}:\mathop {\lim }\limits_{x \to 0} f\left( x \right) = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {x^p}\cos \left( {\frac{1}{x}} \right) = 0 \cr & \Rightarrow \mathop {\lim }\limits_{x \to 0} {x^p}\cos \left( {\frac{1}{x}} \right) = 0 \cr} $$
$$\cos \left( {\frac{1}{x}} \right)$$   is always a finite quantity if $$x \to 0$$
$$ \Rightarrow {x^p} = 0$$
which is possible only if $$p > 0.$$

$${\text{Let }}f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi $$
Doubtful points are $$x = n,\,\,n \in I$$
$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {n - 1} \right)\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr} $$
($$\because \,\left[ x \right]$$ is the greatest integer function)
$$\eqalign{ & {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = \mathop {\lim }\limits_{x \to {n^ + }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = n\,\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr} $$
Now, value of the function at $$x=n$$  is $$f\left( n \right) = 0$$
Since, L.H.L. $$=$$ R.H.L. $$ = f\left( n \right)$$
$$\therefore f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)$$     is continuous for every real $$x.$$

$$t = \frac{1}{{x - 1}}$$   is discontinuous at $$x = 1$$
Also $$y = \frac{1}{{{t^2} + t - 2}}$$    is discontinuous at $$t = - 2$$  and $$t = 1$$
When $$t = - 2,\,\frac{1}{{x - 1}} = - 2 \Rightarrow x = \frac{1}{2}$$
When $$t = 1,\,\frac{1}{{x - 1}} \Rightarrow x = 2$$
So, $$y = f\left( x \right)$$   is discontinuous at three points $$x = 1,\,\frac{1}{2},\,2$$