1. Given $y=\log _{\frac{1}{\sqrt{2}}} \frac{1}{x+3}$, for what value of $x$, (1) $y>0$; (2) $y<0$?
$y=\log _{\frac{1}{\sqrt{2}}} \frac{1}{x+3}=\log _{\sqrt{2}}(x+3)$, so when $y>0$, $x+3>1 \Rightarrow x>-2$; when $y>0$, $x \in(-2,+\infty) ; y<0$ when, $x \in(-3,-2)$.