Three, (Full marks 20 points) In the right triangle $\triangle A B C$, $\angle C=90^{\circ}$. If $\sin A$ and $\sin B$ are the two roots of the equation $x^{2}-\sqrt{2} x-k=0$, find the degree measures of $\angle A$ and $\angle B$ and the value of $k$.

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Three, from the relationship between roots and coefficients, we get
$$
\left\{\begin{array}{l}
\sin A+\sin B=\sqrt{2}, \\
\sin A \sin B=-k .
\end{array}\right.
$$
$$
\begin{array}{l}
\because \angle A+\angle B=90^{\circ}, \\
\therefore \sin B=\cos A .
\end{array}
$$

Thus, we have $\left\{\begin{array}{l}\sin A+\cos A=\sqrt{2}, \\ \sin A \cos A=-k .\end{array}\right.$,
By squaring both sides of (1), we get $\sin A \cos A=\frac{1}{2}$.
From equations (2) and (3), we know $k=-\frac{1}{2}$.
From equations (1) and (3), we know that $\sin A$ and $\cos A$ are the roots of the equation $x^{2}-\sqrt{2} x+\frac{1}{2}=0$, thus $x_{1}=x_{2}=\frac{\sqrt{2}}{2}$.
That is, $\sin A=\cos A=\sin B=\frac{\sqrt{2}}{2}$,
Therefore, $\angle A=\angle B=45^{\circ}$.