Let $$a$$ is a rational number other than 0, in $$\left[ { - 5,\,5} \right],$$  then
$$f\left( a \right) = a\,\,\,{\text{and}}\,\,\,\,\mathop {\lim }\limits_{x \to a} f\left( x \right) = - a$$
[As in the immediate neighborhood of a rational number, we find irrational numbers]
$$\therefore f\left( x \right)$$  is not continuous at any rational number
If $$a$$ is irrational number, then
$$f\left( a \right) = - a\,\,\,{\text{and}}\,\,\,\,\mathop {\lim }\limits_{x \to a} f\left( x \right) = a$$
$$\therefore f\left( x \right)$$  is not continuous at any irrational number clearly
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 0$$
$$\therefore f\left( x \right)$$  is continuous at $$x =0$$

Here, greatest integer function $$\left[ x \right]$$ is discontinuous at its integral value of $$x,\,\cot \,x$$   and $${\text{cosec}}\,x\,$$  are discontinuous at $$0,\,\pi ,\,2\pi $$   etc. and $$\tan \,x$$   and $$sec \,x$$  are discontinuous at $$x = \frac{\pi }{2},\,\frac{{3\pi }}{2},\,\frac{{5\pi }}{2}$$   etc. Therefore the greatest integer function and all trigonometric function are not continuous for $$x\, \in \,R$$
Therefore, neither (1) nor (2) are true.

We have,
$$\eqalign{ & f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\,\sin \,x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\,\cos \,x} \right)}^{2n}}}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\,\sin \,x} \right)}^{2n}}}}{{{{\left( {\sqrt 3 } \right)}^{2n}} - {{\left( {2\,\cos \,x} \right)}^{2n}}}} \cr} $$
$$f\left( x \right)$$  is discontinuous when $${\left( {\sqrt 3 } \right)^{2n}} - {\left( {2\,\cos \,x} \right)^{2n}} = 0$$
$${\text{i}}{\text{.e}}{\text{., }}\cos \,x = \pm \frac{{\sqrt 3 }}{2} \Rightarrow x = n\pi \pm \frac{\pi }{6}\left( {n\, \in \,I} \right)$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^{\frac{1}{x}}} - {e^{ - \frac{1}{x}}}}}{{{e^{\frac{1}{x}}} + {e^{ - \frac{1}{x}}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{1 - {e^{ - \frac{2}{x}}}}}{{1 + {e^{ - \frac{2}{x}}}}} = 1 \cr & {\text{and }}\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{\frac{1}{x}}} - {e^{ - \frac{1}{x}}}}}{{{e^{\frac{1}{x}}} + {e^{ - \frac{1}{x}}}}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{\frac{2}{x}}} - 1}}{{{e^{\frac{2}{x}}} + 1}} = - 1 \cr} $$
Hence, $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$   exists if $$\mathop {\lim }\limits_{x \to 0} g\left( x \right) = 0$$
If $$g\left( x \right) = a \ne 0$$    (constant) then $$\mathop {\lim }\limits_{x \to 0 + } f\left( x \right) = a$$    and $$\mathop {\lim }\limits_{x \to 0 - } f\left( x \right) = - a$$
Thus $$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$   doesn't exist in this case.

$$\eqalign{ & {\text{We have }}f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - \left( x \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( x \right) + f\left( h \right) - f\left( x \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}g\left( h \right)}}{h} \cr & = 0.g\left( 0 \right) \cr & = 0 \cr & \left[ {\because \,g{\text{ is continuous therefore }}\mathop {\lim }\limits_{h \to 0} g\left( h \right) = g\left( 0 \right)} \right] \cr} $$

$$f'\left( x \right) = x\,\sin \,\frac{1}{x}.$$    At all points in $$\left( {0,\,\pi } \right),\,f'\left( x \right)$$   has a definite finite value.
$$\therefore f\left( x \right)$$   is differentiable finitely in $$\left( {0,\,\pi } \right).$$  As a finitely differentiable function is also continuous, we get $$f\left( x \right)$$  is continuous in $$\left( {0,\,\pi } \right).$$

$$\eqalign{ & f\left( x \right) = \left[ {{x^3} - 3} \right] \cr & {\left( {1.26} \right)^3} = 2 - 3 = - 1\,;\,{\left( {1.44} \right)^3} = 3 - 3 = 0 \cr} $$
Similarly, we can check for other points where $$f\left( x \right)$$  changes values to $$1,\,2,\,3,\,4$$
$$\therefore $$  Total number of points of discontinuity are '6'

$$f\left( x \right) = \frac{{1 - \tan \,x}}{{4x - \pi }}$$     is continuous in $$\left[ {0,\,\,\frac{\pi }{2}} \right]$$
$$\eqalign{ & \therefore f\left( {\frac{\pi }{4}} \right) = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} f\left( x \right) = \mathop {\lim }\limits_{x \to \frac{{{\pi ^ + }}}{4}} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {\frac{\pi }{4} + h} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \tan \left( {\frac{\pi }{4} + h} \right)}}{{4\left( {\frac{\pi }{4} + h} \right) - \pi }},\,h > 0 = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \frac{{1 + \tan \,h}}{{1 - \tan \,h}}}}{{4h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2}}{{1 - \tan \,h}}.\frac{{\tan \,h}}{{4h}} \cr & = \frac{{ - 2}}{4} = - \frac{1}{2} \cr} $$

\[\begin{array}{l} f\left( {g\left( x \right)} \right) = f\left( {1 - \left| x \right|} \right) = - 1 + \left| {\left| x \right| + 1} \right|\\ {\rm{Let\,\, }}fog = y\\ \therefore \,y = - 1 + \left| {\left| x \right| + 1} \right| \Rightarrow y\left\{ \begin{array}{l} \,\,\,x,\,\,\,\,x \ge 0\\ - x,\,\,\,\,x < 0 \end{array} \right. \end{array}\]
$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.L}}{\text{. at }}\left( {x = 0} \right) = \mathop {\lim }\limits_{x \to 0} \left( { - x} \right) = 0 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{. at }}\left( {x = 0} \right) = \mathop {\lim }\limits_{x \to 0} \left( x \right) = 0 \cr & {\text{When }}x = 0,{\text{ then }}y = 0 \cr & {\text{Hence, L}}{\text{.H}}{\text{.L}}{\text{. at }}\left( {x = 0} \right) = {\text{ R}}{\text{.H}}{\text{.L at }}\left( {x = 0} \right) = {\text{value of }}y{\text{ at }}\left( {x = 0} \right) \cr} $$
Hence, $$y$$ is continuous at $$x = 0$$
Clearly at all other point $$y$$ continuous. Therefore, the set of all points where $$fog$$  is discontinuous is an empty set.

The function can be continuous only at those points for which $$\sin \,x = \cos \,x \Rightarrow x = n\pi + \frac{\pi }{4}$$