a) Ta có $3\widehat{A1}=2\widehat{A2}$ hay $\widehat{A2}=\frac{3}{2}\widehat{A1}$ và $\widehat{A1}+\widehat{A2}=180^{\circ}$ (hai góc kề bù) nên

$\widehat{A1}+\frac{3}{2}\widehat{A1}=\frac{5}{2}\widehat{A1}=180^{\circ}$ hay $\widehat{A1}=72^{\circ}$. Do đó $\widehat{A2}=\frac{3}{2}72^{\circ}=108^{\circ}$

Từ đó, ta tính được: $\widehat{A1}=\widehat{A3}=\widehat{B1}=\widehat{B3}=72^{\circ}$,$\widehat{A2}=\widehat{A4}=\widehat{B2}=\widehat{B4}=108^{\circ}$