解:$(1)$∵$2²×3²=4×9=36$,$(2×3)²=6²=36$
$(-\frac {1}{2})^2×2^2=\frac {1}{4} ×4=11$,
$[(-\frac {1}{2})×2]^2=(-1)²= 1$
∴每组两个算式的计算结果是相等的
$(2)$根据$(1)$的结果,得$a^nb^n=(\mathrm {ab})^n$
$(3)(\frac {1}{5})^{2023}×(-5)^{2023}$
$=[\frac {1}{5}×(-5)]^{2023}$
$=(-1)^{2023}$
$=-1$