Given that, $2{\log }_{1/4}(x+5)>\frac{9}{4}{\log }_{1/3\sqrt{3}}9+{\log }_{\sqrt{x+5}}2$       ……(i)

By Eq. (i), $x+5>0\Rightarrow x>-5$

$x+5\ne 1\Rightarrow x\ne -4$

So,$x\in (-5,-4)\cup (-4,\infty )$   …(ii)

Now, by Eq. (i)

$\frac{2}{-2}{\log }_2(x+5)-\frac{9}{4}\times \left(\frac{-2}{3}\right){\log }_{3}9-2{\log }_{x+5}2>0$

$-{\log }_2(x+5)-\frac{2}{{\log }_2(x+5)}+3>0$      ……(iii)

Now, let ${\log }_2(x+5)=y,$ then Eq. (iii) becomes

$$\begin{gathered} -y-\frac{2}{y}+3>0 \\ \Rightarrow \frac{-y^2+3y-2}{y}>0 \\ \Rightarrow \frac{y^2-3y+2}{y}<0 \\ \Rightarrow \frac{(y-2)(y-1)}{y}<0 \end{gathered}$$

$$\begin{gathered} \Rightarrow y<0 or 1<y<2 \\ \Rightarrow {\log }_2(x+5)<0 or 1<{\log }_2(x+5)<2 \\ \Rightarrow x+5<1 or 2<x+5<4 \\ \Rightarrow x<-4 ......\left(iv\right) \end{gathered}$$

or $-3<x<-1 ....\left(v\right)$

From Eqs. (ii), (iv) and (v),

$x\in (-5,-4)\cup (-3,-1)$