II. (25 points) Given that $a$ and $b$ are integers, and $a > b$, the equation $3 x^{2} + 3(a+b) x + 4 a b = 0$ has two roots $\alpha, \beta$ that satisfy the relation
$$
\alpha(\alpha+1) + \beta(\beta+1) = (\alpha+1)(\beta+1).
$$

Find all integer pairs $(a, b)$.
From the equation, we get
$$
\alpha+\beta=-(a+b), \alpha \beta=\frac{4}{3} a b \text {. }
$$

From the condition $\alpha(\alpha+1)+\beta(\beta+1)=(\alpha+1)(\beta+1)$, we have $(\alpha+\beta)^{2}-3 \alpha \beta=1$.
Substituting (1) into (2), we get
$$
(a+b)^{2}-4 a b=1 \text {. }
$$

That is, $(a-b)^{2}=1$.
Since $a>b$, we have
$$
a-b=1 \text {. }
$$

From the discriminant $\Delta \geqslant 0$, we get $3(a+b)^{2} \geqslant 16 a b$.
Substituting (3) into (5), we get $(a+b)^{2} \leqslant 4$.
From (5) and (6), we know that the only pairs $(a, b)$ that satisfy the conditions are $(1,0)$ or $(0,-1)$.