Ответ:

Используем правила: \(a^{-n}=\frac{1}{a^n}\), \(a^0=1\).

  1. \[5^{-2}+10^{-3}=\frac{1}{25}+\frac{1}{1000}=\frac{40}{1000}+\frac{1}{1000}=\frac{41}{1000}.\]
  2. \[\left(\frac{6}{7}\right)^{-1}+6^{-2}-(-3,5)^0=\frac{7}{6}+\frac{1}{36}-1=\frac{42}{36}+\frac{1}{36}-\frac{36}{36}=\frac{7}{36}.\]
  3. \[\left(\frac{9}{4}\right)^{-2}\cdot2^{-5}=\left(\frac{4}{9}\right)^2\cdot\frac{1}{2^5}=\frac{16}{81}\cdot\frac{1}{32}=\frac{1}{162}.\]

Ответ: 1) \(\frac{41}{1000}\); 2) \(\frac{7}{36}\); 3) \(\frac{1}{162}\).