Let f x = sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x , for all x ∈ R and g x = π 2…

$\mathrm{f}\left(\mathrm{x}\right)=\sin \left(\frac{\mathrm{\pi }}{6}\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\right)$
$\frac{-\mathrm{\pi }}{2}\le \frac{\mathrm{\pi }}{2}\sin \mathrm{x}\le \frac{\mathrm{\pi }}{2}$
$-1\le \sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\le 1$
$\sin \left(\frac{-\mathrm{\pi }}{6}\right)\le \sin \left(\frac{\mathrm{\pi }}{6}\left(\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\right)\right)\le \sin \frac{\mathrm{\pi }}{6}$
$\frac{-1}{2}\le \sin \left(\frac{\mathrm{\pi }}{6}\left(\sin \left(\frac{\pi }{2}\sin \mathrm{x}\right)\right)\right)\le \frac{1}{2}$
$\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\sin \left(\frac{\mathrm{\pi }}{6}\left(\sin \left(\frac{\mathrm{\pi }}{2}\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\right)\right)\right)$
$-1\le \sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\le 1$
$\frac{-\mathrm{\pi }}{2}\le \frac{\mathrm{\pi }}{2}\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\le \frac{\mathrm{\pi }}{2}$
$-1\le \sin \left(\frac{\mathrm{\pi }}{2}\left(\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\right)\right)\le 1$
$-\frac{\mathrm{\pi }}{6}\le \frac{\mathrm{\pi }}{6}\sin \left(\frac{\mathrm{\pi }}{2}\left(\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\right)\right)\le \frac{\mathrm{\pi }}{6}$
$\frac{-1}{2}\le \sin \left(\frac{\mathrm{\pi }}{6}\sin \left(\frac{\mathrm{\pi }}{2}\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)\right)\right)\le \frac{1}{2}$
$\underset{\mathrm{x}\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\frac{\mathrm{\pi }}{6}\sin \left(\frac{\mathrm{\pi }}{2}\sin x\right)\right)}{\frac{\mathrm{\pi }}{2}\sin \mathrm{x}}\times \frac{\frac{\mathrm{\pi }}{6}\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)}{\frac{\mathrm{\pi }}{6}\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)}$

$=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{\pi }}{6}\frac{\sin \left(\frac{\mathrm{\pi }}{2}\sin \mathrm{x}\right)}{\frac{\mathrm{\pi }}{2}\sin \mathrm{x}}=\frac{\mathrm{\pi }}{6}$.


Range of $\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)$ is $\left[\frac{\mathrm{\pi }}{2}\sin \left(\frac{-1}{2}\right),\frac{\mathrm{\pi }}{2}\sin \frac{1}{2}\right]$
$=\left[-\frac{\mathrm{\pi }}{2}\sin \frac{1}{2},\frac{\mathrm{\pi }}{2}\sin \frac{1}{2}\right]$
Hence, $1$, does not belong to this range.