(2)证明:由题意,得$\overline{aaa1}=1110a+1=3×370a+1。$
$\because 3×370a$是3的倍数,1不是3的倍数,
$\therefore 3×370a+1$不能被3整除。
$\therefore$任意“三联一”数不能被3整除
(3)解:设这两个“三联一”数为$\overline{aaa1},$$\overline{bbb1},$
则$2(\overline{aaa1}+50)+3(\overline{bbb1}+75)=13(171a+256b+25)+2b-3a+5=13k$($k$为整数)。
$\because 1≤ a≤9,$$1≤ b≤4,$且$a,$$b$为整数,$a≠ b。$
$\therefore -20≤2b-3a+5≤10。$
$\therefore 2b-3a+5=-13$或0。
$\therefore 2b-3a=-18$或$-5。$
$\therefore \begin{cases} a=8, \\ b=3 \end{cases}$或$\begin{cases} a=3, \\ b=2 \end{cases}$
$\therefore$这两个数为8881,3331或3331,2221