$$x - \left[ x \right] = 0$$   for all integral values of $$x.$$ Therefore, the function is many-one and, therefore, not defined.

$$\sin \frac{{\pi x}}{{n!}}$$  has the period $$\frac{{2\pi }}{{\frac{\pi }{{n\,!}}}},$$  i.e., $$2\left( {n\,!} \right)$$  and $$\cos \frac{{\pi x}}{{\left( {n + 1} \right)!}}$$   has the period $$\frac{{2\pi }}{{\frac{\pi }{{\left( {n + 1} \right)!}}}},$$   i.e., $$2\left\{ {\left( {n + 1} \right)!} \right\}.$$   Their LCM $$ = 2\left\{ {\left( {n + 1} \right)!} \right\}.$$

The period of $$\sin \frac{{\pi x}}{3}$$   is $$\frac{{2\pi }}{{\frac{\pi }{3}}},$$  i.e., 6. The period of $$\cos \frac{{\pi x}}{4}$$   is $$\frac{{2\pi }}{{\frac{\pi }{4}}},$$  i.e., 8.
LCM of 6 and 8 is 24. So, the period of $$f\left( x \right) = 24.$$

$$\eqalign{ & f\left( {2a - x} \right) = f\left( {a - \left( {x - a} \right)} \right) \cr & = f\left( a \right)f\left( {x - a} \right) - f\left( 0 \right)f\left( x \right) = f\left( a \right)f\left( {x - a} \right) - f\left( x \right) \cr & = - f\left( x \right)\,\left[ {\because x = 0,y = 0,f\left( 0 \right) = {f^2}\left( 0 \right) - {f^2}\left( a \right)} \right. \cr & \left. { \Rightarrow {f^2}\left( a \right) = 0 \Rightarrow f\left( 0 \right) = 0} \right] \cr & \Rightarrow f\left( {2a - x} \right) = - f\left( x \right) \cr} $$

Given that $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \geqslant - 1$$
Clearly $${D_f} = \left[ { - ,\infty } \right)$$   but co-domain is not given. Therefore $$f\left( x \right)$$  need not be necessarily onto.
But if $$f\left( x \right)$$  is onto then as $$f\left( x \right)$$  is one one also, $$\left( {x + 1} \right)$$   being something $$+ve,$$  $${f^{ - 1}}\left( x \right)$$  will exist where $${\left( {x + 1} \right)^2} - 1 = y$$
$$\eqalign{ & \Rightarrow x + 1 = \sqrt {y + 1} \,\left( {{\text{ + ve}}\,{\text{square root as}}\,x + 1 \geqslant 0} \right) \cr & \Rightarrow x = - 1 + \sqrt {y + 1} \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1 \cr & {\text{Then}}\,f\left( x \right) = {f^{ - 1}}\left( x \right) \Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1} - 1 \cr & \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} \Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right) \cr & \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0 \Rightarrow x = - 1,0 \cr} $$
$$\therefore $$ The statement-1 is correct but statement-2 is false.

$$\eqalign{ & {\text{Let}}\,h\left( x \right) = \left| x \right|\,{\text{then}} \cr & g\left( x \right) = \left| {f\left( x \right)} \right| = h\left( {f\left( x \right)} \right) \cr} $$
Since composition of two continuous functions is continuous, therefore $$g$$ is continuous if $$f$$ is continuous.

$$\eqalign{ & {\text{We have, }}F\left( x \right) = \frac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}} \cr & \therefore \,F\left( x \right){\text{ is defined if }}x + 3 > 0{\text{ and }}{x^2} + 3x + 2 \ne 0 \cr & \Rightarrow F\left( x \right){\text{ is defined if }}x > - 3{\text{ and }}x \ne - 1,\, - 2 \cr & \Rightarrow {\text{Domain of }}F\left( x \right) = \left( { - 3,\,\infty } \right) - \left\{ { - 1,\, - 2} \right\} \cr} $$

$${6^x} > 0,\,\,{6^{\left| x \right|}} > 0$$      for all $$x\, \in \,R.$$   So, $$f$$ is into. For different $$x,\,{6^x}$$  and $${6^{\left| x \right|}}$$ are different positive numbers. Clearly, $$f$$ is one-one.

$$\eqalign{ & {\text{Clearly, }}{\left( {0.625} \right)^{4 - 3x}} \geqslant {\left( {1.6} \right)^{x\left( {x + 8} \right)}} \cr & {\text{or }}{\left( {\frac{5}{8}} \right)^{4 - 3x}} \geqslant {\left( {\frac{8}{5}} \right)^{x\left( {x + 8} \right)}} \cr & {\text{or }}{\left( {\frac{8}{5}} \right)^{3x - 4}} \geqslant {\left( {\frac{8}{5}} \right)^{x\left( {x + 8} \right)}} \cr & {\text{or }}3x - 4 \geqslant {x^2} + 8x \cr & {\text{or }}{x^2} + 5x + 4 \leqslant 0 \cr & {\text{or }} - 4 \leqslant x \leqslant - 1 \cr} $$
Hence, the domain of function $$f\left( x \right)$$  is $$x\, \in \left[ { - 4,\, - 1} \right]$$

$$f\left( x \right)$$  is real for those values of $$x$$ for which
$$\eqalign{ & {x^{12}} - {x^9} + {x^4} - x + 1 > 0 \cr & {\text{or }}{x^4}\left( {{x^8} + 1} \right) - x\left( {{x^8} + 1} \right) + 1 > 0 \cr & {\text{or }}\left( {{x^8} + 1} \right)x\left( {{x^3} - 1} \right) + 1 > 0 \cr} $$
If $$x \geqslant 1$$  or $$x \leqslant - 1$$   then the above expression is obviously positive.
If $$ - 1 < x \leqslant 0,$$   the above holds.
If $$0 < x < 1,\,{x^{12}} - x\left( {{x^8} + 1} \right) + \left( {{x^4} + 1} \right) > 0$$         because $${x^4} + 1 > {x^8} + 1$$    and so $${x^4} + 1 > x\left( {{x^8} + 1} \right)$$