Решите пожалуйста,срочно !

1)
$\sqrt[3]{1- \sqrt{2} } \sqrt[6]{3+ 2\sqrt{2} }=\sqrt[6]{(1- \sqrt{2})^2(3+ 2\sqrt{2}) }= \\ =\sqrt[6]{(1- 2\sqrt{2}+2)(3+ 2\sqrt{2})}=\sqrt[6]{(3- 2\sqrt{2})(3+ 2\sqrt{2})}= \\ =\sqrt[6]{9- 4*2}= \sqrt[6]{1}=1$
2)
$(2,5^{ -\frac{2}{9} })^{1,5}( \frac{4}{25} )^{- \frac{1}{6} }=(( \frac{5}{2} )^{ -\frac{2}{9} })^{ \frac{3}{2} }( \frac{2^2}{5^2} )^{- \frac{1}{6} }=( \frac{5}{2} )^{ -\frac{1}{3} }( \frac{2}{5} )^{- \frac{1}{3} }=( \frac{5}{2} * \frac{2}{5} )^{- \frac{1}{3} }=1$

3)
$\sqrt{4+2 \sqrt{3} }- \sqrt{4-2 \sqrt{3} } = (\sqrt{4+2 \sqrt{3} }- \sqrt{4-2 \sqrt{3} })* \frac{\sqrt{4+2 \sqrt{3} }}{\sqrt{4+2 \sqrt{3} }}= \\ = \frac{4+2\sqrt{3}- \sqrt{16-4*3} }{\sqrt{4+2 \sqrt{3} }} = \frac{4+2\sqrt{3}- 2 }{\sqrt{4+2 \sqrt{3} }}=\frac{2+2\sqrt{3}}{\sqrt{4+2 \sqrt{3} }}=\sqrt{ \frac{(2+2 \sqrt{3} )^2}{4+2 \sqrt{3}} }= \\ =\sqrt{ \frac{4+8 \sqrt{3} +12}{4+2 \sqrt{3}} }=\sqrt{ \frac{16+8 \sqrt{3} }{4+2 \sqrt{3}} }=2\sqrt{ \frac{4+2 \sqrt{3} }{4+2 \sqrt{3}} }=2$