a)

Doprowadzamy do potęgi o podstawie $2$:

$\frac{2^{-3}\cdot 4^{-2}}{2^{-6}}=\frac{2^{-3}\cdot {\left(2^2\right)}^{-2}}{2^{-6}}=\frac{2^{-3}\cdot 2^{-4}}{2^{-6}}=$

$=\frac{2^{-7}}{2^{-6}}=2^{-7-\left(-6\right)}=2^{-7+6}=2^{-1}=\frac{1}{2}$

---

b)

Obliczamy:

$\frac{10^{-2}}{5^{-6}\cdot 25^2}=\frac{{\left(2\cdot 5\right)}^{-2}}{5^{-6}\cdot {\left(5^2\right)}^2}=\frac{2^{-2}\cdot 5^{-2}}{5^{-6}\cdot 5^4}=$

$=\frac{2^{-2}\cdot \sout{5^{-2}}}{\sout{5^{-2}}}=2^{-2}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

---

c)

Obliczamy:

$\frac{6^4\cdot 9^{-4}}{4^2\cdot 12^{-1}}=\frac{2^4\cdot 3^4\cdot {\left(3^2\right)}^{-4}}{{\left(2^2\right)}^2\cdot 3^{-1}\cdot 4^{-1}}=\frac{\sout{2^4}\cdot 3^4\cdot 3^{-8}}{\sout{2^4}\cdot 3^{-1}\cdot {\left(2^2\right)}^{-1}}=$

$=\frac{3^{-4}}{3^{-1}\cdot 2^{-2}}=\frac{3^{-3}}{2^{-2}}=\frac{{\left(\frac{1}{3}\right)}^3}{{\left(\frac{1}{2}\right)}^2}=\frac{\frac{1}{27}}{\frac{1}{4}}=\frac{1}{27}\cdot \frac{4}{1}=\frac{4}{27}$

---

d)

Obliczamy:

$\frac{16^{-2}\cdot {125}^{-3}}{10^{-4}\cdot {25}^{-2}}=\frac{{\left(2^4\right)}^{-2}\cdot {\left(5^3\right)}^{-3}}{2^{-4}\cdot 5^{-4}\cdot {\left(5^2\right)}^{-2}}=\frac{2^{-8}\cdot 5^{-9}}{2^{-4}\cdot 5^{-4}\cdot 5^{-4}}=$

$=\frac{2^{-8}\cdot 5^{-9}}{2^{-4}\cdot 5^{-8}}=2^{-4}\cdot 5^{-1}={\left(\frac{1}{2}\right)}^4\cdot \frac{1}{5}=\frac{1}{16}\cdot \frac{1}{5}=\frac{1}{80}$