解:
(1) 解不等式①:
$\begin{aligned}3(x-2)&≤ a-x\\3x+6&≤ a-x\\2x&≤ a-6\\x&≥\frac{6-a}{2}\end{aligned}$
解不等式②:
$\begin{aligned}\frac{2x+1}{3}&≥ x-1\\2x+1&≥3x-3\\x&≥-4\\x&≤4\end{aligned}$
因为不等式组的解集为$2≤ x≤4,$所以$\frac{6-a}{2}=2,$解得$a=2。$
(2) 若不等式组无解,则$\frac{6-a}{2}>4,$
$\begin{aligned}6-a&>8\\a&>2\\a&<-2\end{aligned}$
即$a$的取值范围是$a<-2。$