$\begin{array}{l}
a)3\cot \left( {\dfrac{{3\pi }}{2} - x} \right) + \sqrt 3  = 0\\
 \Leftrightarrow \cot \left( {\dfrac{{3\pi }}{2} - x} \right) =  - \dfrac{{\sqrt 3 }}{3}\\
 \Leftrightarrow \dfrac{{3\pi }}{2} - x =  - \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow  - x =  - \dfrac{\pi }{3} - \dfrac{{3\pi }}{2} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow  - x = \dfrac{{ - 11\pi }}{6} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{{11\pi }}{6} - k\pi \left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{11\pi }}{6} - k\pi \left( {k \in Z} \right)\\
b)\tan 5x.\tan x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan 5x = 0\\
\tan x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
5x = k\pi \\
x = k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{5}\\
x = k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{{k\pi }}{5}\left( {k \in Z} \right)\\
Vậy\,x = \dfrac{{k\pi }}{5}\left( {k \in Z} \right)
\end{array}$