Let $g:R\to R$ be defined as $g\left(x\right)=\frac{x^3+e^{2x}-1}{2}$. If $g\left(f\left(x\right)\right)=xwheref\left(\frac{e^4+7}{2}\right)=\alpha$ , then the number of solution(s) of the equation $||x-2|-3|-\alpha =0$ is (are)

$g\left(2\right)=\frac{e^4+7}{2}$ also g is one-one
Since $g\left(f\left(x\right)\right)=x$  put  $x=\frac{e^4+7}{2}$ 
 $g\left(f\left(\frac{e^4+7}{2}\right)\right)=\frac{e^4+7}{2}=g\left(2\right)$
 $\Rightarrow f\left(\frac{e^4+7}{2}\right)=2\Rightarrow \alpha =2$
From the graph it is clear that equation $||x-2|-3|=2$  has four solutions.