1. (20 points) Given the equation $x^{2}-2 a x+64=0$.
(1) If the equation has real solutions, find the range of real number $a$;
(2) If $a$ is a positive integer, and the equation has integer solutions, find the maximum value of the positive integer $a$.
Three, 1. (1) From the condition, we have
$\Delta=(2 a)^{2}-4 \times 64 \geqslant 0 \Rightarrow a \geqslant 8$ or $a \leqslant-8$.
(2) Let $x_{1}$ be an integer root of the equation $x^{2}-2 a x+64=0$, and the other root be $x_{2}$.

From $x_{1}+x_{2}=2 a$, we know that $x_{2}$ is also an integer root, and $x_{1}$ and $x_{2}$ have the same parity.
By Vieta's formulas, we know $x_{1} x_{2}=64$.
Thus, without loss of generality, assume $x_{2}>x_{1}>0$.
From $(2 a)^{2}=\left(x_{1}+x_{2}\right)^{2}=\left(x_{2}-x_{1}\right)^{2}+4 x_{1} x_{2}$, we know that to maximize $a$, we need $x_{2}-x_{1}$ to be maximized.

From $x_{1} x_{2}=64, x_{2}>x_{1}>0$, and $x_{1}$ and $x_{2}$ have the same parity, then when $x_{2}=32, x_{1}=2$, $a$ is maximized. Hence $a=17$.
Therefore, the maximum value of the positive integer $a$ is 17.