1. (20 points) If positive integers $a$ and $b$ are the roots of the equation $x^{2}-\frac{a^{2}-9}{13} x+10 b+56+5 \sqrt{205+52 b}=0$, find the values of $a$ and $b$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Three, 1. From the conditions, we have
$$
a+b=\frac{a^{2}-9}{13} \text {, }
$$
$$
a b=10 b+56+5 \sqrt{205+52 b} \text {. }
$$

From equation (1), we get $a^{2}-13 a-(13 b+9)=0$.
Since $a$ and $b$ are positive integers, then
$$
a=\frac{13+\sqrt{13^{2}+4(13 b+9)}}{2}=\frac{13+\sqrt{205+52 b}}{2} \text {. }
$$

Therefore, $\sqrt{205+52 b}=2 a-13$.
From equations (2) and (3), we get
$$
a b=10(a+b)-9 \text {. }
$$

Rewriting equation (4) as
$$
(a-10)(b-10)=91 \text {. }
$$

And $91=1 \times 91=7 \times 13=-1 \times(-91)=-7 \times (-13)$. Since $a$ and $b$ are positive integers, $a-10$ and $b-10$ cannot take the values -13 and -91. Therefore, for equation (5), there are only the following four possible cases:
$$
\left\{\begin{array}{l}
a-10=1,91,7,13, \\
b-10=91,1,13,7 .
\end{array}\right.
$$

Solving these, we get $\left\{\begin{array}{l}a=11,101,17,23, \\ b=101,11,23,17 .\end{array}\right.$
Upon verification, only $a=23, b=17$ satisfies equations (1) and (2). In this case, the original equation becomes $x^{2}-40 x+391=0$.
Thus, $a=23, b=17$.