解: (1) 设购买每辆$A$型公交车需要$x$万元,每辆$B$型公交车需要$y$万元,依题意,得$\begin{cases}3x + 2y = 180\\2x + 3y = 195\end{cases},$解得$\begin{cases}x = 30\\y = 45\end{cases}。$
答:购买每辆$A$型公交车需要30万元,每辆$B$型公交车需要45万元。
(2) 设购进$A$型公交车$m$辆,则购进$B$型公交车$(10 - m)$辆,依题意,得$\begin{cases}30m + 45(10 - m) \leq 360\\60m + 100(10 - m) \geq 680\end{cases},$解得$6 \leq m \leq 8。$因为$m$为整数,所以$m = 6,7,8,$所以该公司有三种购车方案,方案1:购进6辆$A$型公交车,4辆$B$型公交车;方案2:购进7辆$A$型公交车,3辆$B$型公交车;方案3:购进8辆$A$型公交车,2辆$B$型公交车。设该公司购买这10辆公交车的总费用为$w$元,则$w = 30m + 45(10 - m)= - 15m + 450,$因为$k = - 15\lt0,$所以$w$随$m$的增大而减小,当$m = 8$时,$w$取得最小值,最小值为330。答:购进8辆$A$型公交车,2辆$B$型公交车时总费用最少,最少总费用为330万元。