(2) Let $a<-1$, and the variable $x$ satisfies $x^{2}+a x \leqslant-x$, and the minimum value of $x^{2}+a x$ is $-\frac{1}{2}$, then $a=$ $\qquad$.
(2) $-\frac{3}{2}$ Hint: From $a-(a+1)$, i.e., $-2<a<-1$, then $f(x)$ takes its minimum value at $x=-(a+1)$, which is $f(-a-1)=a+1$, so $a+1=-\frac{1}{2}, a=-\frac{3}{2}$. If $-\frac{a}{2} \leqslant-(a+1)$, i.e., $a \leqslant-2$, then $f(x)$ takes its minimum value at $x=-\frac{a}{2}$, which is $-\frac{a^{2}}{4}$, so $-\frac{a^{2}}{4}=-\frac{1}{2}, a= \pm \sqrt{2}$ (discard).
In summary, $a=-\frac{3}{2}$.