$解：因為f=100x+42，s=120+y(1≤x≤9，1≤y≤9，且x、y均為整數(shù))$
$f和s的“有趣值”分別記為P_1和 P_2$
$所以P_1=\frac{100x+42-(240+x)}{22}$
$~~~~~~~~~~~~~~=\frac{99x-198}{22}$
$~~~~~~~~~~~~~~=\frac{9(x-2)}{2}$
$P_2=\frac{120+y-(100y+21)}{22}$
$~~~~~~=\frac{99-99y}{22}$
$~~~~~~=\frac{9(1-y)}{2}$
$因為P_1-2P_2=36$
$所以\frac{9(x-2)}{2}-2×\frac{9(1-y)}{2}=36$
$整理可得x+2y=12$
$因為1≤x≤9，1≤y≤9，且x、y均為整數(shù)$
$所以\begin{cases}{x=2}\\{y=5}\end{cases}$
$\begin{cases}{x=4}\\{y=4}\end{cases}$
$\begin{cases}{x=6}\\{y=3}\end{cases}$
$或\begin{cases}{x=8}\\{y=2}\end{cases}$
$將\begin{cases}{x=2}\\{y=5}\end{cases}代入$
$可得P_1=\frac{9(2-2)}{2}=0$
$P_2=\frac{9(1-5)}{2}=-18，符合題意$
$所以\begin{cases}{f=242}\\{s=125}\end{cases}$
$將\begin{cases}{x=4}\\{y=4}\end{cases}代入$
$可得 P_1 =\frac{9(4-2)}{2}=9$
$P_2 =\frac{9(1-4)}{2}=-13.5$
$-13.5不是整數(shù)，不符合題意$
$將\begin{cases}{x=6}\\{y=3}\end{cases}代入$
$可得P_1 =\frac{9(6-2)}{2}=18$
$P_2=\frac{9(1-3)}{2}=-9，符合題意$
$所以\begin{cases}{f=642}\\{s=123}\end{cases}$
$將\begin{cases}{x=8}\\{y=2}\end{cases}代入$
$可得P_1=\frac{9(8-2)}{2}=27$
$P_2=\frac{9(1-2)}{2}=-4.5$
$-4.5不是整數(shù)，不符合題意$
$所以滿足條件的三位數(shù)f 和 s 分別為$
$\begin{cases}{f=242}\\{s=125}\end{cases}$
$\begin{cases}{f=642}\\{s=123}\end{cases}$
