$解:(1)此矩形的周长为(\frac{1}{2}\sqrt{32}+\frac{1}{3}\sqrt{18})×2=(2\sqrt{2}+\sqrt 2)×2=6\sqrt{2}.$
$ \begin{aligned}(2)此矩形的面积&=\frac{1}{2}\sqrt{32}× \frac{1}{3}\sqrt{18} \\ &=2\sqrt{2}× \sqrt{2} \\ &=4. \\ \end{aligned}$
$故与此矩形面积相等的正方形的对角线的长为\sqrt {2^2+2^2} =2\sqrt{2}.$