$a)$    $2^{{\log }_{2}3}=3$  

(zgodnie ze wzorem $a^{{\log }_{a}b}=b$ )

$4^{{\log }_{2}5+{\log }_{\frac{1}{2}}7}=4^{{\log }_{2}5}\cdot 4^{{\log }_{\frac{1}{2}}7}={\left(2^2\right)}^{{\log }_{2}5}\cdot {\left(2^2\right)}^{{\log }_{\frac{1}{2}}7}=2^{2{\log }_{2}5}\cdot 2^{2{\log }_{\frac{1}{2}}7}=$  

$=2^{{\log }_2{5}^2}\cdot {\left(2^{\left(-1\right)\cdot \left(-1\right)}\right)}^{{\log }_{\frac{1}{2}}{7}^2}=2^{{\log }_{2}25}\cdot {\left(2^{-1}\right)}^{\left(-1\right)\cdot {\log }_{\frac{1}{2}}{7}^2}=$

$=25\cdot {\left(\frac{1}{2}\right)}^{{\log }_{\frac{1}{2}}{\left(7^2\right)}^{-1}}=25\cdot {\left(\frac{1}{2}\right)}^{{\log }_{\frac{1}{2}}{7}^{2\cdot \left(-1\right)}}=25\cdot {\left(\frac{1}{2}\right)}^{{\log }_{\frac{1}{2}}{7}^{-2}}=$   

$=25\cdot {\left(\frac{1}{2}\right)}^{{\log }_{\frac{1}{2}}{\left(\frac{1}{7}\right)}^2}=25\cdot {\left(\frac{1}{2}\right)}^{{\log }_{\frac{1}{2}}\left(\frac{1}{49}\right)}=25\cdot \frac{1}{49}=\frac{25}{49}$   

$2^{{\log }_{2}3}=3>\frac{25}{49}=4^{{\log }_{2}5+{\log }_{\frac{1}{2}}7}$  

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$b)$    $\sqrt{0,0007}=\sqrt{\frac{7}{10000}}={\left(\frac{7}{10000}\right)}^{\frac{1}{2}}=\frac{7^{\frac{1}{2}}}{{\left(10000\right)}^{\frac{1}{2}}}=\frac{\sqrt{7}}{\sqrt{10000}}=\frac{\sqrt{7}}{100}\approx \frac{2,65}{100}=$

$=0,026$   

${\left(\frac{1}{3}\right)}^{{\log }_{3}4-{\log }_{\frac{1}{3}}9}={\left(\frac{1}{3}\right)}^{{\log }_{3}4}\cdot {\left(\frac{1}{3}\right)}^{-{\log }_{\frac{1}{3}}9}={\left(3^{-1}\right)}^{{\log }_{3}4}\cdot {\left(\frac{1}{3}\right)}^{{\log }_{\frac{1}{3}}{9}^{-1}}=$  

$=3^{\left(-1\right)\cdot {\log }_{3}4}\cdot {\left(\frac{1}{3}\right)}^{{\log }_{\frac{1}{3}}\left(\frac{1}{9}\right)}=3^{{\log }_3{4}^{-1}}\cdot {\left(\frac{1}{3}\right)}^{{\log }_{\frac{1}{3}}\left(\frac{1}{9}\right)}=3^{\frac{{\log }_{3}1}{4}}\cdot \frac{1}{9}=\frac{1}{4}\cdot \frac{1}{9}=$

$=\frac{1}{36}\approx 0,028$  

$\sqrt{0,0007}\approx 0,026<0,028\approx {\left(\frac{1}{3}\right)}^{{\log }_{3}4-{\log }_{\frac{1}{3}}9}$  

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$c)$    $6^{{\log }_{\frac{1}{6}}\left(\frac{3}{2}\right)+{\log }_{6}9}=6^{{\log }_{\frac{1}{6}}\left(\frac{3}{2}\right)}\cdot 6^{{\log }_{6}9}={\left(6^{\left(-1\right)\cdot \left(-1\right)}\right)}^{{\log }_{\frac{1}{6}}\left(\frac{3}{2}\right)}\cdot 9={\left(6^{-1}\right)}^{\left(-1\right)\cdot {\log }_{\frac{1}{6}}\left(\frac{3}{2}\right)}\cdot 9=$

$={\left(\frac{1}{6}\right)}^{{\log }_{\frac{1}{6}}{\left(\frac{3}{2}\right)}^{-1}}\cdot 9={\left(\frac{1}{6}\right)}^{\frac{{\log }_{\frac{1}{6}}2}{3}}\cdot 9=\frac{2}{3}\cdot 9=\frac{18}{3}=6$   

${\left(\sqrt{7}\right)}^{2{\log }_{7}3+\frac{1}{2}{\log }_{7}16}={\left(\sqrt{7}\right)}^{2{\log }_{7}3+\left(2\cdot \frac{1}{4}\right){\log }_{7}16}={\left(\sqrt{7}\right)}^{2{\log }_{7}3+2\cdot \frac{1}{4}{\log }_{7}16}=$  

$={\left(\sqrt{7}\right)}^{2\cdot \left({\log }_{7}3+\frac{1}{4}{\log }_{7}16\right)}={\left({\left(\sqrt{7}\right)}^2\right)}^{{\log }_{7}3+{\log }_7{16}^{\frac{1}{4}}}=7^{{\log }_{7}3+{\log }_7{16}^{\frac{1}{4}}}=$

$=7^{{\log }_{7}3+{\log }_7{\left(2^4\right)}^{\frac{1}{4}}}=7^{{\log }_{7}3+{\log }_7{2}^{4\cdot \frac{1}{4}}}=7^{{\log }_{7}3+{\log }_{7}2}=7^{{\log }_7\left(3\cdot 2\right)}=$   

$=7^{{\log }_{7}6}=6$   

$6^{{\log }_{\frac{1}{6}}\left(\frac{3}{2}\right)+{\log }_{6}9}=6={\left(\sqrt{7}\right)}^{2{\log }_{7}3+\frac{1}{2}{\log }_{7}16}$  

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$d)$    ${10}^{-1+4\log 5}={10}^{-1}\cdot {10}^{4\log 5}=\left(\frac{1}{10}\right)\cdot {10}^{\log 5^4}=\frac{1}{10}\cdot {10}^{\log 5^4}=\frac{1}{10}\cdot {10}^{\log 625}=$  

$=\frac{1}{10}\cdot 625=\frac{625}{10}=\frac{125}{2}=62\frac{1}{2}$  

$2^{5-\frac{1}{2}{\log }_{2}12}=2^5\cdot 2^{-\frac{1}{2}{\log }_{2}12}=32\cdot 2^{{\log }_2{12}^{-\frac{1}{2}}}=32\cdot 2^{{\log }_2{\left(\frac{1}{12}\right)}^{\frac{1}{2}}}=32\cdot 2^{{\log }_2\sqrt{\frac{1}{12}}}=$  

$=32\cdot 2^{{\log }_2\left(\frac{\sqrt{1}}{\sqrt{12}}\right)}=32\cdot 2^{{\log }_2\left(\frac{1}{\sqrt{4\cdot 3}}\right)}=32\cdot 2^{{\log }_2\left(\frac{1}{\sqrt{4}\cdot \sqrt{3}}\right)}=32\cdot 2^{{\log }_2\left(\frac{1}{2\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\right)}=$

$=32\cdot 2^{{\log }_2\left(\frac{\sqrt{3}}{2\cdot 3}\right)}=32\cdot 2^{{\log }_2\left(\frac{\sqrt{3}}{6}\right)}=32\cdot \frac{\sqrt{3}}{6}=\frac{32\sqrt{3}}{6}=\frac{16\sqrt{3}}{3}\approx$

$\approx \frac{16\cdot 1,73}{3}=\frac{27,68}{3}=9,22\left(6\right)\approx 9$    

${10}^{-1+4\log 5}=62\frac{1}{2}>9\approx 2^{5-\frac{1}{2}{\log }_{2}12}$