1. Given points $A(-1,0), B(1,0), C(0,1)$, the line $y=a x+b \quad(a>0)$ divides triangle $A B C$ into two parts of equal area, then the range of values for $b$ is $\qquad$
1. $\left(1-\frac{\sqrt{2}}{2}, \frac{1}{2}\right)$

Analysis: Case discussion.
(1) When $y=a x+b$ intersects $A C, B C, C O$ at $D, E, F$, $C D \cdot C E=1$,
$\frac{1}{C F}=\left(\frac{1}{C D}+\frac{1}{C E}\right) \sin 45^{\circ}=\frac{\sqrt{2}}{2}\left(C D+\frac{1}{C D}\right)$, and since $C D \in(1, \sqrt{2}]$, we have $C F \in\left[\frac{2}{3}, \frac{\sqrt{2}}{2}\right)$,
thus $b \in\left(1-\frac{\sqrt{2}}{2}, \frac{1}{3}\right]$
(2) When $y=a x+b$ intersects $A B, B C, O C$ at $D, E, F$, let the projection of $E$ on $A B$ be $K, B K=x$, $O K=1-x, B E \cdot B D=\sqrt{2}, B E=\sqrt{2} x, \therefore B D=\frac{1}{x} \in(1,2), \therefore x \in\left(\frac{1}{2}, 1\right), \quad D O=\frac{1}{x}-1$, $\frac{O F}{x}=\frac{\frac{1}{x}-1}{\frac{1}{x}-x}$, we get $O F=\frac{x}{x+1} \in\left(\frac{1}{3}, \frac{1}{2}\right)$, i.e., $b \in\left(\frac{1}{3}, \frac{1}{2}\right)$

In summary, $b \in\left(1-\frac{\sqrt{2}}{2}, \frac{1}{2}\right)$