a) Do $\Delta$ $ABC$ vuông cân tại $A$ nên
$\widehat A=90^o$
$\widehat B=\widehat C=\dfrac{90^o}{2}=45^o$
b) Do $BE\bot EF$
$BC\parallel EF\Rightarrow BC\bot BE$
$\Rightarrow \widehat{EBC}=90^o$
$\Rightarrow \widehat{ABE}=90^o-\widehat{ABC}=90^o-45^o=45^o$
$\widehat{CAF}=\widehat{ACB}=45^o$ (2 góc ở vị trí so le trong)
$\Rightarrow\widehat{ABE}=\widehat{ACB}=45^o$