解$：(1) ∵ $在$△ABC$中，$∠B=66°，∠C=54°， $
$∴ ∠BAC=180°-(∠B+∠C)=180°-(66°+54°)=60°. $
$∵ AD$是$∠BAC$的平分线，
$∴ ∠BAD=∠DAC= \frac {1}{2} ∠BAC=30°$
$∴ ∠ADB=180°-(∠B+∠BAD)=180°-(66°+30°)=84°. $
$∴ ∠ADC=180°-∠ADB=180°-84°=96° $
$(2) ∵ DE⊥AC，$
$∴ ∠DEA=90°. $
$∴ ∠ADE=180°-(∠DAE+∠DEA)=180°-(30°+90°)=60°$
