The Problem: Given: P is the centre of the circle.
CE is the diameter.
∠ 1= 35°
EAB = 100°
Partial Solution: I found the measure of every angle, with the exception of ∠5.
Here's how I found the other angles.
The inscribed angle is half the intercepted arc. So ∠ECB = 1/2 (EAB)
∠EBC = 1/2 (100°) = 50°
EC, as a diameter, cuts the circle in two, making each half 180°.
So EAB + BC = 180°
∠3 is 1/2 (BC) as the inscribed angle is half the intercepted arc.
So ∠3 = 1/2(80°) = 40°
Consider triangle EBC, with interior angles adding up to 180°.
180° -40° (∠3) -50° (∠ECB) -35° (∠1) =∠2
∠2 = 55°
Consider semi-circle EDC.
EP bisects the semicircle in two 90° halves. So DC = 90°
Back to the whole circle, inscribed angle ∠4 is half intercepted arc DC, so ∠4 is 45°.
From here I'm stuck, as I can't find ∠5.
Your question is difficult to understand and read, mainly because of all of the improperly parsed font tags in the original post, but also because it isn't clear which angle you are referring to when you say "∠ = 35°." Fix these problems, and I'm sure someone would be willing to help.
Even though your image was accepted, it is still needlessly large and is inconvenient for people with slower connections.