Verified answer

[tex]\displaystyle\bf\\3)\\\\a^{2} + b^{2} =7ab\\\\\\\frac{2\lg\Big(\dfrac{a+b}{3} \Big)}{\ lga+\ lgb} =\frac{\lg\Big(\dfrac{a+b}{3} \Big)^{2} }{\ lg(ab)} =\frac{\lg\dfrac{a^{2} +b^{2}+2ab }{9} }{\lg(ab)} =\\\\\\=\frac{\lg\dfrac{7ab+2ab }{9} }{\lg(ab)} =\frac{\lg\dfrac{9ab }{9} }{\lg(ab)} =\frac{\lg(ab)}{\lg(ab)} =1[/tex]

[tex]\displaystyle\bf\\4)\\\\13ab=4a^{2}+9b^{2} \\\\\\\frac{2\lg(2a+3b)-2\lg5}{\ lga+\ lgb} =\frac{\lg(2a+3b)^{2} -\lg5^{2} }{\ lg(ab)} =\\\\\\=\frac{\lg\dfrac{(4a^{2}+9b^{2}+12ab) }{25} }{\lg(ab)}= \frac{\lg\dfrac{13ab+12ab }{25} }{\lg(ab)}= \\\\\\= \frac{\lg\dfrac{25ab }{25} }{\lg(ab)}= \frac{\lg(ab)}{\lg(ab)} =1[/tex]

[tex]\displaystyle\bf\\5)\\\\\log_{12} 2=a\\\\\\\log_{6}32=\frac{\log_{12}32}{\log_{12} 6} =\frac{\log_{12} 2^{5} }{\log_{12} \frac{12}{2} } =\frac{5\log_{12}2 }{\log_{12}12-\log_{12} 2 } =\\\\\\=\frac{5\log_{12}2 }{1-\log_{12} 2 } =\frac{5a}{1-a} \\\\6)\\\\\log_{36} 8=a\\\\\log_{36} 2^{3} =a\\\\3\log_{36} 2=a\\\\\boxed{\log_{36} 2=\frac{a}{3}}\\\\\\\log_{36} 9=\log_{36} \frac{36}{4} =\log_{36} 36-\log_{36} 4=1-2\log_{36} 2=\\\\\\=1-2\cdot\frac{a}{3} =\frac{3-2a}{3}[/tex]