Solution set of the inequality $5^{x+2}>\left(\frac{1}{25}\right)^{\frac{1}{x}}$ is:

$(0, \infty)$

We have $5^{x+2}>\left(\frac{1}{25}\right)^{\frac{1}{x}}\left(\begin{array}{l}\text { If}~~a>1 \text {, than } \\ a^m>a^n \Rightarrow m>n\end{array}\right)$

$\Rightarrow 5^{x+2}>5^{-\frac{2}{x}}$

$\Rightarrow x+2>-\frac{2}{x}$

$\Rightarrow\left(\frac{x^2+2 x+2}{x}\right)>0$

as $(x^2+2x+2)=(x+1)^2+1>0$ always

checking for $\frac{1}{x}$ only

$\Rightarrow \frac{1}{x}>0 \Rightarrow x \in(0, \infty)$

Hence (2) is the correct answer.