“测角仪”方案：过$C$作$CF\bot AB$于$F$，

$\because CD\bot BD$，$AB\bot BD$，
$\therefore$四边形$CDBF$是矩形，
$\therefore CF=BD=10m$，$BF=CD=1.6m$，
$\because \angle ACF=32.5^{\circ}$，
$\therefore AF=CF\cdot \tan 32.5^{\circ}=10\times 0.64\approx 6.4\left(m\right)$，
$\therefore AB=AF+BF=6.4+1.6=8\left(m\right)$，
答：树$AB$的高度为$8m$；
“平面镜”方案：$\because CD\bot BD$，$AB\bot BD$，
$\therefore \angle CDE=\angle ABE=90^{\circ}$，
$\because \angle CED=\angle AEB$，
$\therefore \triangle CDE$∽$\triangle ABE$，
$\therefore \frac{CD}{AB}=\frac{DE}{BE}$，
$\therefore \frac{1.6}{AB}=\frac{2}{10}$，
$\therefore AB=8$，
答：树$AB$的高度为$8m$.

“测角仪”方案：过$C$作$CF\bot AB$于$F$，

$\because CD\bot BD$，$AB\bot BD$，
$\therefore$四边形$CDBF$是矩形，
$\therefore CF=BD=10m$，$BF=CD=1.6m$，
$\because \angle ACF=32.5^{\circ}$，
$\therefore AF=CF\cdot \tan 32.5^{\circ}=10\times 0.64\approx 6.4\left(m\right)$，
$\therefore AB=AF+BF=6.4+1.6=8\left(m\right)$，
答：树$AB$的高度为$8m$；
“平面镜”方案：$\because CD\bot BD$，$AB\bot BD$，
$\therefore \angle CDE=\angle ABE=90^{\circ}$，
$\because \angle CED=\angle AEB$，
$\therefore \triangle CDE$∽$\triangle ABE$，
$\therefore \frac{CD}{AB}=\frac{DE}{BE}$，
$\therefore \frac{1.6}{AB}=\frac{2}{10}$，
$\therefore AB=8$，
答：树$AB$的高度为$8m$.