$\text{a)}{\left({\left(6+2\sqrt{2}\right)}^{\frac{\sqrt{2}}{2}}\cdot {\left(6-2\sqrt{2}\right)}^{\frac{\sqrt{2}}{2}}\right)}^{\frac{\sqrt{2}}{2}}={\left\{{\left[\left(6+2\sqrt{2}\right)\left(6-2\sqrt{2}\right)\right]}^{\frac{\sqrt{2}}{2}}\right\}}^{\frac{\sqrt{2}}{2}}={\left(36-8\right)}^{\frac{2}{4}}={28}^{\frac{1}{2}}=\sqrt{28}\notin \mathbf{Q}$  

Podana liczba nie jest wymierna.

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$\text{b)}{\left[{\left(1+2^{\sqrt{2}}\right)}^2-2\left(1+2^{\sqrt{2}}\right)\left(1-2^{\sqrt{2}}\right)+{\left(1-2^{\sqrt{2}}\right)}^2\right]}^{\sqrt{2}-1}={\left\{{\left[\left(1+2^{\sqrt{2}}\right)-\left(1-2^{\sqrt{2}}\right)\right]}^2\right\}}^{\sqrt{2}-1}={\left[{\left(1+2^{\sqrt{2}}-1+2^{\sqrt{2}}\right)}^2\right]}^{\sqrt{2}-1}={\left[{\left(2\cdot 2^{\sqrt{2}}\right)}^2\right]}^{\sqrt{2}-1}={\left[{\left(2^{\sqrt{2}+1}\right)}^2\right]}^{\sqrt{2}-1}=2^{2\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2^{2\cdot \left(2-1\right)}=2^2=4\in \mathbf{Q}$  

Podana liczba jest wymierna.