a. AC is a diameter of the circle, so arc ABDC has measure 180º. Then
[tex]135^\circ+(2t)^\circ+t^\circ=180^\circ[/tex]
[tex]\implies135+3t=180[/tex]
[tex]\implies3t=45[/tex]
[tex]\implies\boxed{t=15}[/tex]
b. The angle formed by the two secants AE and BE has measure equal to half the difference between the intercepted arcs:
[tex]m\angle AEB=\dfrac{m\widehat{AB}-m\widehat{CD}}2[/tex]
[tex]\implies m\angle AEB=\dfrac{135^\circ-15^\circ}2=\boxed{120^\circ}[/tex]
