Five. (Full marks 20 points, (1) 12 points, (2) 8 points)
If the sines of the two acute angles $A, B$ of a right triangle are the two roots of the equation $x^{2}+p x+q=0$:
(1) What conditions should the real numbers $p, q$ satisfy?
(2) If $p, q$ satisfy these conditions, are the two roots of the equation $x^{2}+p x+q=0$ equal to the sines of the two acute angles $A, B$ of the right triangle?
$$
\begin{array}{l}
(1) \text{ Let } A, B \text{ be the two acute angles of a right-angled triangle, and } \sin A, \sin B \text{ are the roots of the equation } x^{2}+p x+q=0, \text{ then } \\
\Delta=p^{2}-4 q \geqslant 0 . \\
\sin A+\sin B=-p, \sin A \sin B=q . \\
\because \sin A>0, \sin B>0, \\
\therefore p<0 . \\
\because \sin B=\cos A . \\
\therefore \sin A+\cos A=-p, \sin A \cos A=q, \\
\therefore p^{2}=(-p)^{2}=(\sin A+\cos A)^{2} \\
\quad=1+2 \sin A \cos A=1+2 q,
\end{array}
$$

So, $p^{2}-2 q=1$.
From (1), $1-2 q=p^{2}-4 q \geqslant 0$, then $q \leqslant \frac{1}{2}$.
Therefore, the required condition is
$$
p<0 .
\end{array}
$$
$$
\begin{array}{l}
\because \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=p^{2}-2 q=1, \\
\therefore 0<\alpha \leqslant \beta<1 .
\end{array}
$$
$\therefore \alpha, \beta$ are the sines of the two acute angles of a right-angled triangle.
$$