By mean value theorem, there exists a real number $$c\, \in \left( {2,\,4} \right)$$   such that
$$\eqalign{ & f'\left( c \right) = \frac{{f\left( 4 \right) - f\left( 2 \right)}}{{4 - 2}}\,\,\, \Rightarrow f'\left( c \right) = \frac{{f\left( 4 \right) + 4}}{2} \cr & {\text{Since, }}f'\left( c \right) \geqslant 6,\,\forall \,x\, \in \left[ {2,\,4} \right] \cr & \therefore \,f'\left( c \right) \geqslant 6 \cr & \Rightarrow \frac{{f\left( 4 \right) + 4}}{2} \geqslant 6 \cr & \Rightarrow f\left( 4 \right) + 4 \geqslant 12 \cr & \Rightarrow f\left( 4 \right) \geqslant 8 \cr} $$

$$\eqalign{ & {\text{Given, }}f\left( x \right) = \frac{1}{x} - \frac{2}{{{e^{2x}} - 1}} \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{1}{x} - \frac{2}{{{e^{2x}} - 1}} \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{e^{2x}} - 1} \right) - 2x}}{{x\left( {{e^{2x}} - 1} \right)}}\,\,\,\left[ {\frac{0}{0}\,\,{\text{from}}} \right] \cr} $$
$$\therefore $$ using, L'Hospital rule
$$\eqalign{ & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{4{e^{2x}}}}{{2\left( {x{e^{2x}}\,2 + {e^{2x}}.1} \right) + {e^{2x}}.2}} \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{4{e^{2x}}}}{{4x{e^{2x}}\, + 2{e^{2x}} + 2{e^{2x}}}}\,\,\,\left[ {\frac{0}{0}\,\,{\text{from}}} \right] \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{4{e^{2x}}}}{{4\left( {x{e^{2x}}\, + {e^{2x}}} \right)}}\,\, = \frac{{4.{e^0}}}{{4\left( {0 + {e^0}} \right)}}\,\, = 1 \cr} $$

$$f\left( x \right)$$  is continuous at $$x=1$$  if
$$\eqalign{ & \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = f\left( 1 \right) = 2 \cr & \mathop {\lim }\limits_{h \to 0} f\left( {1 + h} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {1 + h + 1} \right\} = 2 \cr & \mathop {\lim }\limits_{h \to 0} f\left( {1 - h} \right) = \mathop {\lim }\limits_{h \to 0} \left\{ {p{{\left( {1 - h} \right)}^2} - q} \right\} = p - q \cr} $$
$$\therefore \,f\left( x \right)$$   is not continuous at $$x=1$$  if $$p - q \ne 2$$

We have $$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$$
$$ = \mathop {\lim }\limits_{h \to 0} \sin \left( {{{\log }_e}\left| { - h} \right|} \right)$$
$$ = \mathop {\lim }\limits_{h \to 0} \sin \left( {{{\log }_e}h} \right)$$    which does not but lies between $$-1$$ and $$1$$
Similarly, $$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$$   lies between $$-1$$ and $$1$$ but cannot be determined.

$$\eqalign{ & \mathop {{\text{Lt}}}\limits_{x \to 0} \frac{{{{\left( {{e^x} - 1} \right)}^2}}}{{\sin \left( {\frac{x}{a}} \right)\log \left( {1 + \frac{x}{4}} \right)}} = \mathop {{\text{Lt}}}\limits_{x \to 0} \frac{{\frac{{{{\left( {{e^x} - 1} \right)}^2}}}{x}.{x^2}}}{{\frac{x}{a}.\frac{{\sin \left( {\frac{x}{a}} \right)}}{{\left( {\frac{x}{a}} \right)}}.\frac{{\log \left( {1 + \frac{x}{4}} \right)}}{{\left( {\frac{x}{4}} \right)}}.\frac{x}{4}}} \cr & \Rightarrow 4a = 12 \cr & \Rightarrow a = 3 \cr} $$

As $$f$$ is continuous,
$$\eqalign{ & f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right) \cr & = \mathop {\lim }\limits_{n \to \infty } f\left( {\frac{1}{{4n}}} \right) \cr & = \mathop {\lim }\limits_{n \to \infty } \left( {\left( {\sin \,{e^n}} \right){e^{ - {n^2}}} + \frac{1}{{1 + \frac{1}{{{n^2}}}}}} \right) \cr & = 0 + 1 \cr & = 1 \cr} $$

Given, $$f\left( {x\,y} \right) = f\left( x \right).\,f\left( y \right)$$     for all $$x,y$$ .....(1)
$$f\left( x \right)$$  is continuous at $$x = 2,$$  i.e., $$\mathop {{\text{Lt}}}\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)$$    .....(2)
$$\eqalign{ & {\text{Let }}a \ne 0 \cr & {\text{Now, }}\mathop {{\text{Lt}}}\limits_{x \to a} f\left( x \right) = \mathop {{\text{Lt}}}\limits_{h \to 2} f\left( {\frac{{ah}}{2}} \right) \cr & \left[ {{\text{putting }}x = \frac{{ah}}{2}{\text{ so that }}h = \frac{{2x}}{a}} \right] \cr & = f\left( {\frac{a}{2}} \right)\mathop {{\text{Lt}}}\limits_{h \to 2} f\left( h \right) \cr & = f\left( {\frac{a}{2}} \right).f\left( 2 \right) \cr & = f\left( {\frac{a}{2}.2} \right) \cr & = f\left( a \right) \cr} $$
Hence, $$f\left( x \right)$$  is necessarily continuous at $$x = a$$  for all $$a \ne 0.$$
At $$x = 0,\,f\left( x \right)$$   may or may not be continuous.
Hence, $$f\left( x \right)$$  is not necessarily continuous in $$\left( { - \infty ,\, + \infty } \right)$$

$$\eqalign{ & {\text{Let }}x = a\, \in \,Q\,;\,f\left( a \right) = a \cr & f\left( {{a^ + }} \right) = 1 - a{\text{ or }}a\,;\,f\left( {{a^ - }} \right) = 1 - a{\text{ or }}a\, \cr & {\text{continuous at where }}1 - a = a \Rightarrow a = \frac{1}{2} \cr & \Rightarrow {\text{continuous at one point}}. \cr} $$