7.013. $\frac{\left(25^{\frac{1}{2 \log _{4} 25}}+2 \log _{2} \log _{2} \log _{2} a^{2 \log _{a} 4}\right) \cdot 4^{-\frac{2}{\log _{3} 4}}-a^{2}}{1-a}$.
## Solution.

$$
\frac{\left(25^{\frac{1}{2 \log _{9} 25}}+2 \log _{2} \log _{2} \log _{2} a^{2 \log _{a} 4}\right) \cdot 4^{\frac{2}{\log _{3} 4}}-a^{2}}{1-a}=
$$

$$
\begin{aligned}
& =\frac{\left(\left(25^{\log _{2} 49}\right)^{\frac{1}{2}}+2 \log _{2} \log _{2} 4\right) \cdot\left(\left(4^{\log _{4} 3}\right)^{-1}-a^{2}\right)}{1-a}= \\
& =\frac{\left((49)^{\frac{1}{2}}+2 \log _{2} 2\right) \cdot 9^{-1}-a^{2}}{1-a}=\frac{(7+2) \cdot \frac{1}{9}-a^{2}}{1-a}=\frac{1-a^{2}}{1-a}=1+a .
\end{aligned}
$$

Answer: $1+a$.