a. m∠AOD = m arc AD = 35
b. DO bisects ∠AOB, so m∠DOB = m∠AOD = 35 = m arc DB
c. m∠AOD + m∠DOB + m∠BOC = 180
35 + 35 + m∠BOC = 180 → m∠BOC = 180 - 35 - 35 = 110 = m arc BC
d. m∠A + 90 + m∠AOD = 180
m∠A + 90 + 35 = 180 → m∠A = 180 - 90 - 35 = 55
e. m∠C = (1/2) * m arc BA
m∠C = (1/2) * (35 + 35) = 35
f. Since m∠AOD = m∠C , DE is || to BC... and so m∠AFO = m∠B = 90
g. m∠COE = m∠AOD = 35 = m arc CE
h. m arc AE = 180 - m arc CE = 180 - 35 = 145
Looks like my answers agree with yours!
Also I checked your number 8 and I agree with all your answers for it! 

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