解：（1）因为$BD，$$CE$都是$\triangle ABC$的角平分线，
所以$\angle DBC=\angle ABD=\frac{1}{2}\angle ABC，$$\angle ECB=\angle ACE=\frac{1}{2}\angle ACB。$
则$\angle DBC+\angle ECB=\frac{1}{2}(\angle ABC+\angle ACB)=\frac{1}{2}(180^{\circ}-\angle A)=90^{\circ}-\frac{1}{2}\angle A。$
所以$\angle BOC = 180^{\circ}-(\angle DBC+\angle ECB)=180^{\circ}-(90^{\circ}-\frac{1}{2}\angle A)=90^{\circ}+\frac{1}{2}\angle A。$
又因为$\angle BOC-\angle A = 54^{\circ}，$即$90^{\circ}+\frac{1}{2}\angle A-\angle A = 54^{\circ}，$
$90^{\circ}- \frac{1}{2}\angle A=54^{\circ}，$
$\frac{1}{2}\angle A=90^{\circ}-54^{\circ}，$
$\frac{1}{2}\angle A = 36^{\circ}，$
$\angle A = 72^{\circ}。$
所以$\angle BOC = 90^{\circ}+\frac{1}{2}\times72^{\circ}=90^{\circ}+36^{\circ}=126^{\circ}。$
（2）因为$BD，$$CE$都是$\triangle ABC$的高，所以$\angle ADB=\angle AEC = 90^{\circ}。$
因为$\angle A+\angle ADB+\angle DOE+\angle AEC = 360^{\circ}，$
所以$\angle A + 90^{\circ}+\angle DOE+90^{\circ}=360^{\circ}，$$\angle A=180^{\circ}-\angle DOE。$
又因为$\angle DOE=\angle BOC，$所以$\angle A = 180^{\circ}-\angle BOC。$
因为$\angle BOC-\angle A = 54^{\circ}，$所以$\angle BOC-(180^{\circ}-\angle BOC)=54^{\circ}，$
$2\angle BOC=54^{\circ}+180^{\circ}，$
$2\angle BOC = 234^{\circ}，$
$\angle BOC = 117^{\circ}。$
（3）$\angle ODC-\angle BEO = 18^{\circ}。$
理由：因为$\angle BEO=\angle A+\angle ACE，$
$\angle BOC=\angle BEO+\angle ABD=\angle A+\angle ACE+\angle ABD，$
所以$\angle BOC-\angle A=\angle ACE+\angle ABD。$
因为$\angle BOC-\angle A = 54^{\circ}，$$\angle ABD = 2\angle ACE，$
所以$54^{\circ}=\angle ACE+2\angle ACE，$
$3\angle ACE = 54^{\circ}，$
$\angle ACE = 18^{\circ}。$
所以$\angle ABD = 2\times18^{\circ}=36^{\circ}。$
因为$\angle BOC=\angle ODC+\angle DCO=\angle BEO+\angle ABD，$
所以$\angle BEO + 36^{\circ}=\angle ODC+18^{\circ}，$
则$\angle ODC-\angle BEO = 18^{\circ}。$
