答案：
4.(1)证明:$\because$ 在$△ AOB$中,$∠ A+∠ B+∠ AOB=180°$,
在$△ COD$中,$∠ C+∠ D+∠ COD=180°$,
$\therefore ∠ A+∠ B+∠ AOB=∠ C+∠ D+∠ COD.$
又$\because ∠ AOB=∠ COD,\therefore ∠ A+∠ B=∠ C+∠ D.$
(2)解:如答图①,$\because AP,CP$分别平分$∠ BAD,∠ BCD$,
$\therefore ∠ 1=∠ 2,∠ 3=∠ 4.$

由(1)的结论,得$\begin{cases}∠ P+∠ 3=∠ 2+∠ B,①\\∠ P+∠ 1=∠ 4+∠ D,②\end{cases}$
①+②,得$2∠ P+∠ 1+∠ 3=∠ 2+∠ 4+∠ B+∠ D$,
即$2∠ P=∠ 2+∠ 4+∠ B+∠ D-∠ 1-∠ 3$,
$\therefore 2∠ P=∠ B+∠ D.$
$\because ∠ ABC=36°,∠ ADC=16°$,
$\therefore 2∠ P=52°,\therefore ∠ P=26°.$
(3)解:如答图②.
$\because AM$平分$∠ FAD,CP$平分$∠ BCE$,
$\therefore ∠ 1=∠ 2,∠ 3=∠ 4$,
$\therefore ∠ PAD=180°-∠ 2,∠ PCD=180°-∠ 3.$
$\because$ 由(1)知$∠ P+∠ PAD=∠ D+∠ PCD$,
$\therefore ∠ P+(180°-∠ 2)=∠ D+(180°-∠ 3).$
$\because ∠ P+∠ PAB=∠ B+∠ 4,∠ 1=∠ PAB$,
$\therefore 2∠ P=∠ B+∠ D. \because ∠ ABC=30°,∠ ADC=10°$,
$\therefore ∠ P=\frac{1}{2}(∠ ABC+∠ ADC)=\frac{1}{2}×(30°+10°)=20°.$

答案：
5.解:(1)$\because CE$平分$∠ BCD$,$AE$平分$∠ BAD$,
$\therefore ∠ ECD = ∠ ECB = \frac{1}{2}∠ BCD, ∠ EAD = ∠ EAB = \frac{1}{2}∠ BAD.$
$\because ∠ D+∠ ECD=∠ E+∠ EAD, ∠ B+∠ EAB=∠ E+∠ ECB,$
$\therefore ∠ D+∠ ECD+∠ B+∠ EAB=∠ E+∠ EAD+∠ E+∠ ECB,$
$\therefore ∠ D+∠ B=2∠ E,\therefore ∠ E=\frac{1}{2}(∠ D+∠ B).$
$\because ∠ D=40°,∠ B=30°,\therefore ∠ E=\frac{1}{2}×(40°+30°)=35°.$
(2)与(1)同理,可得$∠ E=\frac{1}{2}(∠ D+∠ B)$,
$\because ∠ D=m°,∠ B=n°,\therefore ∠ E=\frac{m°+n°}{2}.$
(3)存在.$∠ E=\frac{∠ B-∠ D}{2}.$
证明:延长 BC 交 AD 于点 F,如答图.

$\because ∠ BFD=∠ B+∠ BAD,$
$\therefore ∠ BCD=∠ BFD+∠ D=∠ B+∠ BAD+∠ D.$
$\because CE$平分$∠ BCD$,$AE$平分$∠ BAD$,$\therefore ∠ ECD=∠ ECB=\frac{1}{2}∠ BCD,∠ EAD=∠ EAB=\frac{1}{2}∠ BAD.$
$\because ∠ E+∠ ECB=∠ B+∠ EAB,$
$\therefore ∠ E=∠ B+∠ EAB-∠ ECB=∠ B+∠ EAB-\frac{1}{2}∠ BCD=∠ B+∠ EAB-\frac{1}{2}(∠ B+∠ BAD+∠ D)=\frac{1}{2}(∠ B-∠ D),即∠ E=\frac{∠ B-∠ D}{2}.$