(12 分) (1) 已知关于 $x$ 的实系数方程 $x^{2}+m x+n=0$, 若 $1+\sqrt{2 i}$ 是方程
(1) 由题得,$$\left\{egin{array} { l } { - 1 + m + n = 0 } \{ 2 \sqrt { 2 } + m \sqrt { 2 } = 0 }\end{array} , 	ext { 解得 } \left\{egin{array}{l}m=-2 \n=3\end{array}
ight.
ight.$$(2) 设 $z=x+y i(x, y \in \mathbf{R}), ecause z+3 i=x+(y+3) \mathrm{i}$ 为实数, $	herefore y=-3$.$ecause rac{z}{3-\mathrm{i}}=rac{x-3 \mathrm{i}}{3-\mathrm{i}}=rac{1}{10}(x-3 \mathrm{i})(3+\mathrm{i})-rac{1}{10}[(3 x+3)+(x-9) \mathrm{i}]$ 为实数, $	herefore x=9, 	herefore z=9-3 \mathrm{i}$.$ecause(z+a i)^{2}=81-(a-3)^{2}+18(a-3) i=72+6 a-a^{2}+18(a-3) i$, $	herefore$ 由已知得 $\left\{egin{array}{l}72+6 a-a^{2}>0 \ 18(a-3)>0\end{array}
ight.$, 解得 $3<a<12$, 即 $a$ 的取值范围是 $(3,12)$.