a)

$\sqrt{25\cdot 121\cdot 4}=\sqrt{25}\cdot \sqrt{121}\cdot \sqrt{4}=5\cdot 11\cdot 2=110$

b)

$\sqrt{\frac{{\sout{9}}^1}{{\sout{16}}^4}\cdot \frac{{\sout{4}}^1}{{\sout{81}}^9}}=\sqrt{\frac{1}{36}}=\frac{\sqrt{1}}{\sqrt{36}}=\frac{1}{6}$