Solution: 1. A set is closed if and only if its complement is open. So we list the complement of each sets in $\mathcal{T} $, and they will be all closed subsets of $X$? \[ \emptyset,X,\{b,c,d,e\}, \{c,d,e\}, \{b,c,e\}, \{e\}, \{c,d\}, \{b,e\}. \]

2. Recall that for any set $A$, the closure is defined as the intersection of all closed sets containing $A$, that is, \[ \bar{A} \coloneqq \bigcap \left\{ C\supseteq A: C \text{ is closed in }X \right\}. \] So we have \begin{align*} \overline{\{a\}} & = X \\ \overline{\{b,c\}} & = X\cap\{b,c,d,e,\}\cap \{b,c,e\} \\ & = \{b,c,e\}\\ \overline{\{d\}} & = X\cap\{b,c,d,e,\}\cap \{c,d,e\} \cap \{c,d\}\\ & = \{c,d\} \\ \overline{\{a,e\}} & = X \\ \overline{\{a,c,d\}} & = X. \end{align*}

3. Recall that a set $A$ is dense in $X$ if and only if $\bar{A}=X$. So, the dense subsets are \[ \{a\}, \{a,e\},\{a,c,d\}. \]