9. (16 points) Given the parabola $y=x^{2}+p x+q$ passes through the points $\left(x_{1}, 0\right),\left(x_{2}, 0\right),(1, a),(2, b)$, where $1<x_{1}<x_{2}<2$.
(1) If $x_{1}=\frac{7}{5}, x_{2}=\frac{8}{5}$, find the values of $a$ and $b$;
(2) Prove: If $a \geqslant \frac{1}{4}$, then $b<\frac{1}{4}$.
(1) From the condition, we have
$$
x^{2}+p x+q=\left(x-x_{1}\right)\left(x-x_{2}\right) \text {. }
$$

Substituting \( x=1 \) and \( x=2 \), and combining with \( 10, \\
b=4+2 p+q=\left(2-x_{1}\right)\left(2-x_{2}\right)>0 .
\end{array}
$$

Taking \( x_{1}=\frac{7}{5}, x_{2}=\frac{8}{5} \), substituting into equations (1) and (2) respectively, we get
$$
a=\frac{6}{25}, b=\frac{6}{25} \text {. }
$$
(2) Multiplying (1) by (2) gives
$$
\begin{array}{l}
a b=\left(1-x_{1}\right)\left(1-x_{2}\right)\left(2-x_{1}\right)\left(2-x_{2}\right) \\
=\left(\left(1-x_{1}\right)\left(2-x_{1}\right)\right)\left(\left(1-x_{2}\right)\left(2-x_{2}\right)\right) \\
=\left(\frac{1}{4}-\left(x_{1}-\frac{3}{2}\right)^{2}\right)\left(\frac{1}{4}-\left(x_{2}-\frac{3}{2}\right)^{2}\right) .
\end{array}
$$

Given \( 1<x_{1}<x_{2}<2 \), we know
$$
\begin{array}{l}
0<\frac{1}{4}-\left(x_{1}-\frac{3}{2}\right)^{2} \leqslant \frac{1}{4}, \\
0<\frac{1}{4}-\left(x_{2}-\frac{3}{2}\right)^{2} \leqslant \frac{1}{4} .
\end{array}
$$

Thus, \( 0<a b \leqslant \frac{1}{4^{2}} \), and when and only when \( x_{1}=x_{2}=\frac{3}{2} \),
$$
a b=\frac{1}{4^{2}} \text {. }
$$

Since \( x_{1} \neq x_{2} \), we have \( 0<a b<\frac{1}{4^{2}} \).
Therefore, when \( a \geqslant \frac{1}{4} \), \( b<\frac{1}{4} \).