3. Find all pairs of natural numbers $a$ and $b$, for which out of the four statements:
1) $a^{2}+4 a+3$ is divisible by $b$;
2) $a^{2}+a b-6 b^{2}-2 a-16 b-8=0$;
3) $a+2 b+1$ is divisible by 4;
4) $a+6 b+1$ is a prime number

three are true, and one is false.
Solution. Statements 3) and 4) cannot both be true. If 3) is true, then $a+6b+1=(a+2b+1)+4b$ is divisible by 4 and is not a prime number. Therefore, one of the statements 3) or 4) is false.

Let's consider statement 2$): a^{2}+a(b-2)-6b^{2}-16b-8=0$. Consider this equation as a quadratic in $a$. Calculate the discriminant $D=(b-2)^{2}+24b^{2}+64b+32=(5b+6)^{2}$. We have $a=2b+4$ or $a=-3b-2$. The latter equality does not hold since $a$ and $b$ are natural numbers, $a>0, -3b-2<0$. Statements 2) and 3) cannot both be true. If 2) is true, then substituting $a=2b+4$ into 3) gives $a+2b+1=4b+5$, which is not divisible by 4. Therefore, one of the statements 2) or 3) is false. Considering that there is only one false statement among the four, the false statement can only be 3).

Thus, for natural numbers $a$ and $b$, the following conditions are satisfied: $a^{2}+4a+3$ is divisible by $b$; $a=2b+4$; $a+6b+1$ is a prime number, i.e., $(2b+4)^{2}+4(2b+4)+3$ is divisible by $b$, and $8b+5$ is a prime number. We have $(2b+4)^{2}+4(2b+4)+3=4b^{2}+24b+35$. If this number is divisible by $b$, then $b$ can be $1, 5, 7$, or 35.

1) $b=1, a=6, 8b+5=13$ - a prime number, $b=1, a=6$ is a solution to the problem.
2) $b=5, a=14, 8b+5=45$ - not a prime number, this case does not provide a solution.
3) $b=7, a=18, 8b+5=61$ - a prime number, $b=7, a=18$ is a solution to the problem.
4) $b=35, a=74, 8b+5=5 \cdot 57$ - not a prime number, this case does not provide a solution.

Answer: $a=6, b=1; a=18, b=7$.