1. If the quadratic equation with integer coefficients
$$
x^{2}+(a+3) x+2 a+3=0
$$

has one positive root $x_{1}$ and one negative root $x_{2}$, and $\left|x_{1}\right|<\left|x_{2}\right|$, then
$$
a=
$$
$\qquad$
$=、 1 .-2$.
Since the two roots of the equation are not equal, we have $\Delta>0$, that is
$$
(a+3)^{2}>4(2 a+3) \text {. }
$$

Solving this, we get $a>3$ or $a-3, a<-\frac{3}{2}$, which means $-3<a<-\frac{3}{2}$.
Since $a$ is an integer, then $a=-2$.