Let us analyze the transitions of the given DFA M:
State 1 is both the start state and the only accepting state: F={1}.
δ(1,0)=2, δ(1,1)=4
δ(2,0)=3, δ(2,1)=1
δ(3,0)=1, δ(3,1)=2
δ(4,0)=1, δ(4,1)=5
δ(5,0)=4, δ(5,1)=1
### 1. State Distinguishability / Equivalence
Pair (2,5):
On input 0: δ(2,0)=3 and δ(5,0)=4.
On input 1: δ(2,1)=1∈F and δ(5,1)=1∈F.
Pair (3,4):
On input 0: δ(3,0)=1∈F and δ(4,0)=1∈F.
On input 1: δ(3,1)=2 and δ(4,1)=5.
Since both transitions from {2,5} lead to {3,4} on 0 and {1} on 1, and both transitions from {3,4} lead to {1} on 0 and {2,5} on 1, states 2 and 5 are indistinguishable (2≡5), and states 3 and 4 are indistinguishable (3≡4).
Statement (B) says states 3 and 4 are distinguishable, which is FALSE (they are equivalent).
Statement (C) says states 2 and 5 are distinguishable, which is FALSE (they are equivalent).
### 2. Pair (2,4)
On input 1: δ(2,1)=1∈F, but δ(4,1)=5∈/F.
Thus, string 1 distinguishes states 2 and 4. Statement (A) is TRUE.
### 3. Language of the DFA
The states track the value of (n0(w)−n1(w))(mod3):
State 1: (n0(w)−n1(w))≡0(mod3)
States {2,5}: (n0(w)−n1(w))≡1(mod3)
States {3,4}: (n0(w)−n1(w))≡2(mod3)
If n0(w)=n1(w), then n0(w)−n1(w)=0≡0(mod3), which leads to state 1 (the accepting state). Therefore, any such string is accepted. Statement (D) is TRUE.
Since the question asks for the FALSE statements, the correct options are B and C.