解：$(1) $由题意，得
$F(243)=(423+342+234)÷111=9，$$F(617)=(167+716+671)÷111=14$
$(2) $由题意，得$F(\mathrm {s})=(302+10x+230+x+100x+23)÷111=x+5，$
$F(\mathrm {t})=(510+y+100y+51+105+10y)÷111=y+6$
∴当$F(\mathrm {s})+F(\mathrm {t})=18$时，$x+5+y+6=18$
∴$x+y=7$
∵$s，$$t $都是“相异数”
∴$x≠2，$$x≠3$且$y≠1，$$y≠5$
∵$1≤x≤9，$$1≤y≤9，$且$x，$$ y $都是正整数
∴$\{\begin{array}{l}x=1\\y=6\end{array} $或$ \{\begin{array}{l}x=4\\y=3\end{array} $或$ \{\begin{array}{l}x=5\\y=2\end{array} $
∵$k=\frac {F(\mathrm {s})}{F(\mathrm {t})}$
∴分类讨论如下:
①当$ \{\begin{array}{l}x=1\\y=6\end{array} $时，$ F(\mathrm {s})=6，$$ F(\mathrm {t})=12$
∴$k=\frac {1}{2}$
②当$ \{\begin{array}{l}x=4\\y=3\end{array} $时，$ F(\mathrm {s})=9，$$ F(\mathrm {t})=9$
∴$k=1$
③当$ \{\begin{array}{l}x=5\\y=2\end{array} $时，$ F(\mathrm {s})=10，$$ F(\mathrm {t})=8$
∴$k= \frac {5}{4}$
∵$\frac {1}{2}<1<\frac {5}{4}$
∴$k $的最大值为$ \frac {5}{4}$
