Đặt \({\rm{lo}}{{\rm{g}}_9}a = {\rm{lo}}{{\rm{g}}_{16}}b = {\rm{lo}}{{\rm{g}}_{12}}\frac{{5b - a}}{2} = t\).

\( \Leftrightarrow a = {9^t};b = {16^t};\frac{{5b - a}}{2} = {12^t}\)

\(\frac{{5b - a}}{2} = {12^t} \Leftrightarrow 5b - a = {12^t}.2\)

\( \Leftrightarrow {5.16^t} - {9^t} - {2.12^t} = 0\)

\( \Leftrightarrow 5.{\left( {{4^t}} \right)^2} - {2.4^t}{.3^t} - {\left( {{3^t}} \right)^2} = 0\)

\( \Leftrightarrow 5.{\left[ {{{\left( {\frac{4}{3}} \right)}^t}} \right]^2} - 2.{\left( {\frac{4}{3}} \right)^t} - 1 = 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{{{\left( {\frac{4}{3}} \right)}^t} = \frac{{1 + \sqrt 6 }}{5}\left( {TM} \right)}\\{{{\left( {\frac{4}{3}} \right)}^t} = \frac{{1 - \sqrt 6 }}{5}{\rm{\;(KTM)\;}}}\end{array}} \right.\)

\( \Leftrightarrow {\left( {\frac{4}{3}} \right)^t} = \frac{{1 + \sqrt 6 }}{5} \Rightarrow t = {\rm{lo}}{{\rm{g}}_{\frac{4}{3}}}\frac{{1 + \sqrt 6 }}{5} < 0\).

\(\frac{a}{b} = {\left[ {{{\left( {\frac{3}{4}} \right)}^t}} \right]^2} = {\left( {\frac{5}{{1 + \sqrt 6 }}} \right)^2}\).

\( \Rightarrow \) Khẳng định 1 và 3 sai, khẳng định 2 đúng.

Do đó ta chọn đáp án như sau