Đáp án:

$\begin{array}{l}
1)\sqrt {3a} .\sqrt {27a} \left( {a \ge 0} \right)\\
 = \sqrt {81{a^2}} \left( {a \ge 0} \right)\\
 = 9.a\\
2)\sqrt {3{a^3}} .\sqrt {12a} \left( {a > 0} \right)\\
 = \sqrt {36{a^4}} \\
 = 6{a^2}\\
3)\sqrt {2a.32a{b^2}} \left( {a > 0;b > 0} \right)\\
 = \sqrt {64{a^2}{b^2}} \left( {a > 0;b > 0} \right)\\
 = 8ab\\
4)\sqrt {27.48{{\left( {1 - a} \right)}^2}} \left( {a > 1} \right)\\
 = \sqrt {81.16{{\left( {1 - a} \right)}^2}} \\
 = 9.4.\left( {a - 1} \right)\\
 = 36\left( {a - 1} \right)\\
5)\sqrt {{a^4}{{\left( {3 - a} \right)}^2}} \left( {a \ge 3} \right)\\
 = {a^2}.\left( {a - 3} \right)\\
 = {a^3} - 3{a^2}\\
6)\dfrac{1}{{a - b}}.\sqrt {{a^4}.{{\left( {a - b} \right)}^2}} \left( {a > b} \right)\\
 = \dfrac{1}{{a - b}}.{a^2}.\left( {a - b} \right)\\
 = {a^2}'\\
7)\sqrt {\dfrac{{4{a^2}}}{{25}}} \left( {a > 0} \right)\\
 = \dfrac{{2a}}{5}\\
8)\dfrac{{\sqrt {27a} }}{{\sqrt {3a} }} = \dfrac{{3\sqrt 3 .\sqrt a }}{{\sqrt 3 .\sqrt a }} = 3\\
9)\sqrt {\dfrac{{2{a^2}{b^4}}}{{50}}}  = \dfrac{{\sqrt {{a^2}{b^4}} }}{{\sqrt {25} }} = \dfrac{{\left| a \right|{b^2}}}{5}\\
10)\dfrac{{2a{b^2}}}{{\sqrt {162} }} = \dfrac{{2a{b^2}}}{{9\sqrt 2 }} = \dfrac{{2\sqrt 2 a{b^2}}}{{18}}\\
11)\sqrt {\dfrac{{27{{\left( {a - 3} \right)}^2}}}{{48}}}  = \sqrt {\dfrac{{9{{\left( {a - 3} \right)}^2}}}{{16}}}  = \dfrac{{3\left( {a - 3} \right)}}{4}\\
\left( {a > 3} \right)\\
12)\sqrt {\dfrac{{9 + 12a + 4{a^2}}}{{{b^2}}}} \\
 = \dfrac{{\sqrt {{{\left( {2a + 3} \right)}^2}} }}{{\sqrt {{b^2}} }} = \dfrac{{2a + 3}}{{ - b}} =  - \dfrac{{2a + 3}}{b}\\
13)\left( {a - b} \right).\sqrt {\dfrac{{ab}}{{{{\left( {a - b} \right)}^2}}}} \left( {a < b < 0} \right)\\
 = \left( {a - b} \right).\dfrac{{\sqrt {ab} }}{{b - a}}\\
 =  - \sqrt {ab} 
\end{array}$