$解：(1)如图所示$
$\text{(2)作}AM⊥BC\text{，垂足为点}M.$
$设BM=x\,\,\text{m}，则CM=\text{(}6-x\text{)\ m}$
$在Rt△ABM中，由勾股定理可知，$
$AM^2＋BM^2=AB^2∵AB=4\ \text{m}，BM=x\ \text{m}$
$∴AM^2=AB^2-BM^2=16-x^2$
$在Rt△ACM中，由勾股定理可知，$
$AM^2+CM^2=AC^2$
$∵CM=\left( 6-x \right) \ \text{m,\ }AC=5\ \text{m}$
$∴16-x^2+\left( 6-x \right) ^2=5^2解得x=\frac{9}{4}$
$∴BM=\frac{9}{4}\ \text{m,\ }AM=\frac{5\sqrt{7}}{4}\ \text{m}$
$∴S_{△ABC}=\frac{1}{2}×6×\frac{5\sqrt{7}}{4}=\frac{15\sqrt{7}}{4}\ \text\ \mathrm {{m}^2}$
$设所画圆形花坛的半径为\ r\ \text{m}，$
$则\frac{1}{2}×\text{(}4+5+6\text{)}r=\frac{15\sqrt{7}}{4}解得，r=\frac{\sqrt{7}}{2}$
$∴所画圆形花坛的半径为\frac{\sqrt{7}}{2}\ \text{m}.$
