解：$(1) $正确$.$设这两个$“$好数$”$为$a + b\sqrt {2}$，
$c + d\sqrt {2}(a$，$b$，$c$，$d$为有理数$)$，
则$(a + b\sqrt {2})(c + d\sqrt {2}) $
$= ac + ad\sqrt {2} + bc\sqrt {2} + bd·2$
$=(ac + 2bd)+(ad + bc)\sqrt {2}$，
其中$ac + 2bd$为有理数，$ad + bc $为有理数，所以这两个“好数”
之积为“好数”.
$(2) $正确$.$设这两个好数为$a + b\sqrt {2}$，$c + d\sqrt {2}(a$，$b$，$c$，$d$为有理数，$cd≠0)$，
则$\frac {a + b\sqrt {2}}{c + d\sqrt {2}}$
$=\frac {(a + b\sqrt {2})(c - d\sqrt {2})}{(c + d\sqrt {2})(c - d\sqrt {2})}$
$=\frac {(ac + 2bd)+(ad + bc)\sqrt {2}}{c^2-2d^2}$
$=\frac {ac + 2bd}{c^2-2d^2}+\frac {ad + bc}{c^2-2d^2}\sqrt {2}$，
其中$\frac {ac + 2bd}{c^2-2d^2}$，$\frac {ad + bc}{c^2-2d^2}$均为有理数，
所以这两个“好数”之商为“好数”.
