Question 1209678
Since $f(x)$ is divisible by $x^2-3$, the roots of $x^2-3=0$ must also be roots of $f(x)$. The roots of $x^2-3=0$ are $x=\sqrt{3}$ and $x=-\sqrt{3}$.

Therefore, $f(\sqrt{3})=0$ and $f(-\sqrt{3})=0$.

$f(\sqrt{3}) = (\sqrt{3})^3 + a(\sqrt{3})^2 + (b+2)\sqrt{3} + 1 = 0$
$3\sqrt{3} + 3a + (b+2)\sqrt{3} + 1 = 0$
$(3+b+2)\sqrt{3} + 3a + 1 = 0$
$(b+5)\sqrt{3} + 3a + 1 = 0$

$f(-\sqrt{3}) = (-\sqrt{3})^3 + a(-\sqrt{3})^2 + (b+2)(-\sqrt{3}) + 1 = 0$
$-3\sqrt{3} + 3a - (b+2)\sqrt{3} + 1 = 0$
$-(b+5)\sqrt{3} + 3a + 1 = 0$

We have a system of two equations:
1) $(b+5)\sqrt{3} + 3a + 1 = 0$
2) $-(b+5)\sqrt{3} + 3a + 1 = 0$

Adding the two equations gives:
$2(3a+1) = 0$
$3a+1 = 0$
$3a = -1$
$a = -\frac{1}{3}$

Substituting $a=-\frac{1}{3}$ into the first equation:
$(b+5)\sqrt{3} + 3(-\frac{1}{3}) + 1 = 0$
$(b+5)\sqrt{3} - 1 + 1 = 0$
$(b+5)\sqrt{3} = 0$
$b+5 = 0$
$b = -5$

Thus, the ordered pair of constants is $(-\frac{1}{3}, -5)$.

Final Answer: The final answer is $\boxed{\left(-\frac{1}{3}, -5\right)}$