a)

$\frac{{\left(1,4\right)}^{-5}\cdot {\left(\frac{5}{7}\right)}^{-5}+5^{\frac{1}{2}}\cdot 5^{\frac{1}{2}}+7^{\frac{1}{2}}\cdot 7^{\frac{1}{2}}}{\left(3^{1,5}-\sqrt{7}\right)\left(3^{1,5}+\sqrt{7}\right)}=\frac{{\left(1,4\cdot \frac{5}{7}\right)}^{-5}+5^{\frac{1}{2}+\frac{1}{2}}+7^{\frac{1}{2}+\frac{1}{2}}}{{\left(3^{1,5}\right)}^2-{\sqrt{7}}^2}=$      

$=\frac{{\left(\frac{14}{10}\cdot \frac{5}{7}\right)}^{-5}+5^1+7^1}{3^3-7}=\frac{1^{-5}+5+7}{27-7}=\frac{1+5+7}{20}=\frac{13}{20}$  

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b)

$\frac{\left(3^{\frac{3}{2}}-\sqrt{3}\right)\left(3^{\frac{3}{2}}+\sqrt{3}\right)}{{\left(0,75\right)}^{-3}\cdot {\left(1\frac{1}{3}\right)}^{-3}+{23}^{\frac{1}{3}}\cdot {23}^{\frac{1}{3}}\cdot {23}^{\frac{1}{3}}}=\frac{{\left(3^{\frac{3}{2}}\right)}^2-{\sqrt{3}}^2}{{\left(0,75\cdot 1\frac{1}{3}\right)}^{-3}+{23}^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}=$  

$=\frac{3^3-3}{{\left(\frac{3}{4}\cdot \frac{4}{3}\right)}^{-3}+{23}^1}=\frac{27-3}{1^{-3}+23}=\frac{24}{1+23}=\frac{24}{24}=1$  

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c)

$\frac{\left(6-{11}^{\frac{1}{2}}\right)\left(6+{11}^{\frac{1}{2}}\right)}{{\left(\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}\right)}^2}=\frac{6^2-{\left({11}^{\frac{1}{2}}\right)}^2}{{\sqrt{6-\sqrt{11}}}^2-2\cdot \sqrt{6-\sqrt{11}}\sqrt{6+\sqrt{11}}+{\sqrt{6+\sqrt{11}}}^2}=$  

$=\frac{36-{11}^1}{6-\sqrt{11}-2\cdot \sqrt{\left(6-\sqrt{11}\right)\left(6+\sqrt{11}\right)}+6+\sqrt{11}}=\frac{36-11}{12-2\sqrt{6^2-{\sqrt{11}}^2}}=$  

$=\frac{25}{12-2\sqrt{36-11}}=\frac{25}{12-2\sqrt{25}}=\frac{25}{12-2\cdot 5}=\frac{25}{12-10}=\frac{25}{2}=12\frac{1}{2}$  

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d)

${\left[{\left(3-\sqrt{5}\right)}^{\frac{1}{2}}+{\left(3+\sqrt{5}\right)}^{\frac{1}{2}}\right]}^{-2}\cdot \left[{\left(\frac{27}{8}\right)}^{\frac{1}{3}}-{28}^0\right]=$  

$=\left[\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right){]}^{-2}\cdot \left[{\left({\left(\frac{3}{2}\right)}^3\right)}^{\frac{1}{3}}-1\right]=$  

$=\frac{1}{{\left[\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right]}^2}\cdot \left[{\left(\frac{3}{2}\right)}^1-1\right]=$  

$=\frac{1}{{\sqrt{3-\sqrt{5}}}^2+2\sqrt{3-\sqrt{5}}\sqrt{3+\sqrt{5}}}+{\sqrt{3+\sqrt{5}}}^2]\cdot \left[\frac{3}{2}-1\right]=$  

$=\frac{1}{3-\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+3+\sqrt{5}}\cdot \frac{1}{2}=$  

$=\frac{1}{6+2\sqrt{3^2-{\sqrt{5}}^2}}\cdot \frac{1}{2}=\frac{1}{6+2\sqrt{9-5}}\cdot \frac{1}{2}=$  

$=\frac{1}{6+2\sqrt{4}}\cdot \frac{1}{2}=\frac{1}{6+2\cdot 2}\cdot \frac{1}{2}=\frac{1}{6+4}\cdot \frac{1}{2}=\frac{1}{10}\cdot \frac{1}{2}=\frac{1}{20}$