By definition, we know that:
[tex]\begin{gathered} D\cup E=\mleft\lbrace w\mright|w\in D\text{ or }w\in E\} \\ =\lbrace w|w\le2\text{ or }w<5\} \end{gathered}[/tex]
Since 2<5, then the only way w can be out of this union is if 5≤w. Then:
[tex]\begin{gathered} D\cup E=\mleft\lbrace w\mright|w<5\} \\ =(-\infty,5) \end{gathered}[/tex]
On the other hand:
[tex]\begin{gathered} D\cap E=\mleft\lbrace w\mright|w\in D\text{ and }w\in E\} \\ =\lbrace w|w\le2\text{ and }w<5\} \end{gathered}[/tex]
Since 2<5, then w is in the intersection only if w≤2. Then:
[tex]\begin{gathered} D\cap E=\lbrace w|w\le2\} \\ =(-\infty,2\rbrack \end{gathered}[/tex]
Therefore, the answers are:
[tex]\begin{gathered} D\cup E=(-\infty,5) \\ D\cap E=(-\infty,2\rbrack \end{gathered}[/tex]
