To determine which statements are true about the finite automata \(D_1\) and \(D_2\), we first need to analyze the languages accepted by each automaton.

Step 1: Analyze \(D_1\) and \(D_2\)

Finite automata \(D_1\) and \(D_2\) are defined over the alphabet \(\{0, 1\}\). Each automaton transitions based on input characters. The task is to evaluate the operations and find out about:

1. The equivalence of languages: \(L(D_1) = L(D_2)\)
2. Subset relation: \(L(D_1) \subset L(D_2)\)
3. Intersection: \(L(D_1) \cap L(D_2) = \{\epsilon\}\)
4. Closure property: \((L(D_1) \cup L(D_2))^*\) resulting in strings whose length is divisible by 3

Step 2: Determine \(L(D_1)\) and \(L(D_2)\)

Without explicit transition details, determining the exact languages requires observation or given transition rules. Normally, this involves enumerating or deducing from patterns what each automaton accepts.

Step 3: Evaluate Statements

• \(L(D_1) = L(D_2)\): This would imply both automata accept exactly the same set of strings. The provided solution does not confirm this.
• \(L(D_1) \subset L(D_2)\): This would suggest that all strings accepted by \(D_1\) are also accepted by \(D_2\) but not vice versa. This is unconfirmed without specific analysis.
• \(L(D_1) \cap L(D_2) = \{\epsilon\}\): This states only the empty string \(\epsilon\) is in both languages, meaning they share no other common strings. This is marked as correct in the information given.
• \((L(D_1) \cup L(D_2))^*\) consists of all strings whose length is divisible by 3: This claims that the repeated union (closure) of both languages leads to strings with lengths divisible by 3. This also is marked as correct.

Conclusion:

• The correct statements are: \(L(D_1) \cap L(D_2) = \{\epsilon\}\) and \((L(D_1) \cup L(D_2))^*\) consists of all strings whose length is divisible by 3.