$\text{a)}\frac{2}{3^{0,5}-1}=\frac{2}{\sqrt{3}-1}=\frac{2}{\sqrt{3}-1}\cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\left(\sqrt{3}+1\right)}{{\left(\sqrt{3}\right)}^2-1^2}=\frac{2\left(\sqrt{3}+1\right)}{3-1}=\frac{2\left(\sqrt{3}+1\right)}{2}=\sqrt{3}+1$