$解:由(3)知$
$a2+5ab+6b2=(a+3b)(a+2b)$
$∴(a+3b)(a+2b)=35$
$∵a,b都是正整數(shù)$
$∴\begin{cases}{a+3b=35}\\{a+2b=1}\end{cases}$
$或\begin{cases}{a+3b=7}\\{a+2b=5}\end{cases}$
$解得\begin{cases}{a=-67}\\{b=34}\end{cases}(舍去)$
$或\begin{cases}{a=1}\\{b=2}\end{cases}$
$答:a的值為1,b的值為2.$