Step 1: Recall the AND gate rule.
A two-input AND gate outputs $Y = A \cdot B$. The output is $1$ only when both inputs are $1$; in every other case the output is $0$.
Step 2: Test entry 1.
With $A = 0, B = 1$: $Y = 0 \cdot 1 = 0$. The listed value matches, so entry 1 is correct.
Step 3: Test entry 2.
With $A = 1, B = 0$: $Y = 1 \cdot 0 = 0$. Correct, so entry 2 is valid.
Step 4: Test entry 3.
With $A = 1, B = 1$: $Y = 1 \cdot 1 = 1$. Correct, so entry 3 is valid.
Step 5: Test entry 4.
With $A = 0, B = 0$: the true output is $Y = 0$, but the table shows $Y = 1$. This is wrong, so entry 4 is invalid.
Step 6: Collect the correct entries.
Entries 1, 2 and 3 are correct while 4 is not, so the answer is 1, 2 and 3 only, option (2).
\[ \boxed{\text{1, 2 and 3 only}} \]