Solution.

$$
\begin{aligned}
& \left(81^{\frac{1}{4}-\frac{1}{2} \log _{9} 4}+25^{\log _{125} 8}\right) \cdot 49^{\log _{7} 2}=\left(\frac{81^{\frac{1}{4}}}{\left(9^{2}\right)^{\frac{1}{2} \log _{9} 4}}+5^{2 \log _{5} 32^{3}}\right) \cdot 7^{2 \log _{7} 2}= \\
& =\left(\frac{3}{4}+4\right) \cdot 4=19
\end{aligned}
$$

Answer: 19.
7.007. $\left(81^{\frac{1}{4}-\frac{1}{\log _{9} 4}}+25^{\log _{125} 8}\right) \cdot 49^{\log _{7} 2}$.