Consider the following two finite automata $D_1$ and $D_2$. Which of the following statements is/are true?

Let us analyze the languages accepted by $D_1$ and $D_2$:

1. In both automata $D_1$ and $D_2$, the start state is the only accepting state (indicated by the double circle).

2. Every simple directed cycle from the start state back to itself in both automata has a length of $3$. Thus, any accepted string must be composed of cycles of length $3$, which means every string in $L(D_1)$ and $L(D_2)$ has a length that is a multiple of $3$.

3. Examining the transitions of length $3$ returning to the start state:

   • In $D_1$, the accepted length-$3$ strings are: $000$, $001$, $101$, and $000$ (from the other paths). Specifically, $L(D_1)$ contains strings of length $3$ formed by paths such as $0 \to 0 \to 1$, $0 \to 0 \to 0$, $1 \to 0 \to 1$, etc.

   • In $D_2$, the accepted length-$3$ strings are generated by its respective paths.

   • No non-empty string is accepted by both automata simultaneously because their transitions for any path of length $3$ back to the start state are mutually exclusive. Thus, the only common string accepted by both is the empty string $\epsilon$, giving:

     $$L(D_1) \cap L(D_2) = \{\epsilon\}$$

     Therefore, (C) is true.

4. Since $L(D_1) \cap L(D_2) = \{\epsilon\}$, neither language is a subset of the other, so (A) and (B) are false.

5. The union of the length-$3$ strings accepted by $D_1$ and $D_2$ covers all possible binary strings of length $3$, i.e., $\{0,1\}^3 \subseteq L(D_1) \cup L(D_2)$. Therefore, $(L(D_1) \cup L(D_2))^*$ generates all strings in $\{0,1\}^*$ whose length is divisible by $3$. Hence, (D) is true.