$解:(1)過點(diǎn)D作DF⊥BC于點(diǎn)F.$
$根據(jù)題意，得DF=2\sqrt{3}m，EF=2m.$
$∵BE=4m，$
$∴在Rt△DFB中,tan∠ABC=\frac{DF}{BF}=\frac{2\sqrt{3}}{2+4}=\frac{\sqrt{3}}{3}.$
$∴∠ABC=30°$
$(2)如圖, 過點(diǎn)A作AH⊥BP于點(diǎn)H$
$∵∠ACP=2∠ABC=60°$
$∴∠BAC=∠ABC=30°$
$∴AC=BC=BE+EC=8m$
$在Rt△ACH中，AH=AC \cdot sin ∠ACP=8×\frac {\sqrt{3}}2=4\sqrt{3}(\mathrm {m})$
$∴光源所在的點(diǎn)A處距水平面的高度為4\sqrt{3}m$
