解：（1）$\angle BDC=\angle BAC+\angle B+\angle C。$理由：
如图，连接$AD$并延长至点$F。$
因为$\angle FDC = \angle DAC + \angle C，$$\angle BDF = \angle B + \angle BAD，$
所以$\angle FDC+\angle BDF=\angle DAC+\angle C+\angle B+\angle BAD，$
即$\angle BDC = \angle BAC + \angle B + \angle C。$
（2）①因为$\angle BDC = 90^{\circ}，$
由（1）知$\angle A+\angle ABD+\angle ACD=\angle BDC = 90^{\circ}。$
因为$\angle A = 40^{\circ}，$
所以$\angle ABD + \angle ACD = 90^{\circ}-40^{\circ}=50^{\circ}。$
②因为$\angle BPC = 130^{\circ}，$$\angle A = 40^{\circ}，$
由（1）可知，
$\angle ABP + \angle ACP = \angle BPC - \angle A = 130^{\circ}-40^{\circ}=90^{\circ}。$
因为$BD$平分$\angle ABP，$$CD$平分$\angle ACP，$
所以$\angle ABD=\frac{1}{2}\angle ABP，$$\angle ACD=\frac{1}{2}\angle ACP。$
所以$\angle ABD + \angle ACD=\frac{1}{2}\angle ABP+\frac{1}{2}\angle ACP=\frac{1}{2}(\angle ABP+\angle ACP)=45^{\circ}。$
由（1）可知，
$\angle BDC = \angle A + \angle ABD + \angle ACD = 40^{\circ}+45^{\circ}=85^{\circ}。$
