$\text{a)}$  

$2^{5\sqrt{5}}\cdot {32}^{2-\sqrt{5}}=2^{5\sqrt{5}}\cdot {\left(2^5\right)}^{2-\sqrt{5}}=2^{5\sqrt{5}}\cdot 2^{10-5\sqrt{5}}=2^{5\sqrt{5}+10-5\sqrt{5}}=2^{10}=2^{2\cdot 5}={\left(2^2\right)}^5=4^5$  

Zatem:

$2^{5\sqrt{5}}\cdot {32}^{2-\sqrt{5}}=4^5$  

$\text{b)}$  

$9^{\frac{\sqrt{7}}{2}}\cdot {27}^{-\frac{\sqrt{7}}{3}}={\left(3^2\right)}^{\frac{\sqrt{7}}{2}}\cdot {\left(3^3\right)}^{-\frac{\sqrt{7}}{3}}=3^{\sqrt{7}}\cdot 3^{-\sqrt{7}}=3^{\sqrt{7}-\sqrt{7}}=3^0=1$