(1) Given the sets $A=\left\{y \mid y=x^{2}+2 x-3\right\}, B=\left\{y \left\lvert\, y=x+\frac{1}{x}\right., x<\right.$ 0), then $A \cap B=$ . $\qquad$
(1) From $y=(x+1)^{2}-4 \geqslant-4$, we get $A=[-4,+\infty)$. Since $x<0$, $y=$ $x+\frac{1}{x} \leqslant-2$, so $B=(-\infty,-2]$. Therefore, $A \cap B=[-4,-2]$.