1. Given $\sin x+\sin y=\frac{1}{3}$. Then the maximum value of $\sin y-\cos ^{2} x$ is . . $\qquad$
$$
\text { Ni. } 1 \cdot \frac{4}{9} \text {. }
$$

From $\sin y=\frac{1}{3}-\sin x$ belonging to $[-1,1]$, we know
$$
-\frac{2}{3} \leqslant \sin x \leqslant \frac{4}{3} \text {. }
$$

Thus, $-\frac{2}{3} \leqslant \sin x \leqslant 1$.
$$
\begin{aligned}
\text { Also, } & \sin y-\cos ^{2} x \\
& =\left(\frac{1}{3}-\sin x\right)-\left(1-\sin ^{2} x\right) \\
& =\sin ^{2} x-\sin x-\frac{2}{3} \\
& =\left(\sin x-\frac{1}{2}\right)^{2}-\frac{11}{12} .
\end{aligned}
$$

When $\sin x=-\frac{2}{3}$, it takes the maximum value $\frac{4}{9}$.