III/1. (a) First, we simplify the exponent:

$a=\frac{\log _{7} 4\left(\log _{7} 5-\log _{7} 2\right)}{\log _{7} 25\left(\log _{7} 8-\log _{7} 4\right)}=\frac{\log _{7} 2^{2} \cdot \log _{7} \frac{5}{2}}{\log _{7} 5^{2} \cdot \log _{7} 2}=\frac{2 \cdot \log _{7} 2 \cdot \log _{7} \frac{5}{2}}{2 \cdot \log _{7} 5 \cdot \log _{7} 2}=\frac{\log _{7} \frac{5}{2}}{\log _{7} 5}$.

The last expression is rewritten in base 5: $\frac{\frac{\log _{5} \frac{5}{2}}{\log _{5} 7}}{\log _{5} 7}=\log _{5} \frac{5}{2}$. Finally, we have: $5^{a}=5^{\log _{5} \frac{5}{2}}=\frac{5}{2}$.

(b) From the solution (a), we see that the result is independent of the base of the logarithm. Therefore, $5^{b}=$ $5^{a}=\frac{5}{2}$.

Calculation of $a=\log _{5} \frac{5}{2}: 3$ points. Conclusion $5^{a}=\frac{5}{2}: \mathbf{2}$ points. Calculation $5^{b}=\frac{5}{2}: \mathbf{2}$ points.

If the contestant correctly solves only (b) with the appropriate calculation $b=\log _{5} \frac{5}{2}$ and the conclusion $5^{b}=\frac{5}{2}: 3+2$ points.
1. (a) What is $5^{a}$, if $a=\frac{\log _{7} 4\left(\log _{7} 5-\log _{7} 2\right)}{\log _{7} 25\left(\log _{7} 8-\log _{7} 4\right)}$?

(b) What is $5^{b}$, if $b=\frac{\log _{77} 4\left(\log _{77} 5-\log _{77} 2\right)}{\log _{77} 25\left(\log _{77} 8-\log _{77} 4\right)}$?