$\begin{array}{l} D = \left\{ {x \in \mathbb{Z}|\left( {x - 2} \right)\left( {4{x^2} + 5x + 1} \right) = 0} \right\}\\ \left( {x - 2} \right)\left( {4{x^2} + 5x + 1} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right)\left( {4x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = - 1 \end{array} \right.\left( {do\,x \in \mathbb{Z}} \right)\\ \Rightarrow D = \left\{ {2; - 1} \right\}\\ E = \left\{ {x \in \mathbb{R}|{x^2} + \left( {3m + 1} \right)x + 3m = 0} \right\}\\ {x^2} + \left( {3m + 1} \right)x + 3m = 0\\ a = 1,b = 3m + 1,c = 3m \Rightarrow a - b + c = 0\\ \Rightarrow x = - 1,x = - 3m\\ \Rightarrow D = \left\{ { - 1; - 3m} \right\}\\ \Rightarrow D \cup E = \left\{ {2; - 1; - 3m} \right\}\\ \left( {m \ne \dfrac{1}{3}do\, - 3m \ne 2; - 1} \right)\\ \Rightarrow {2^2} + {\left( { - 1} \right)^2} + {\left( { - 3m} \right)^2} = 6\\ \Leftrightarrow 4 + 1 + 9{m^2} = 6\\ \Leftrightarrow 9{m^2} = 1 \Leftrightarrow {m^2} = \dfrac{1}{9}\\ \Leftrightarrow m = \pm \dfrac{1}{3} \Rightarrow m = - \dfrac{1}{3} \end{array}$
