This is a really nice number-theoretic combinatorial geometry with an algebra touch, therefore this is a combinatorics problem.
a) Answer. $\boxed{\text{No.}}$ Let us consider cyclic $19$-gon $A_0A_1A_2\ldots A_{18}$ with circumcentre $O$. Thus, since all its angles are integers, we must have that $$\frac{\angle{A_iOA_{i+1}}+\angle{A_{i+1}OA_{i+2}}}{2}\in\mathbb N$$for every $i=0,1,\ldots,18$, where the indices are taken modulo $19$. Since our polygon has odd number of vertices, then we must have that $\angle{A_iOA_{i+1}}\equiv \angle{A_jOA_{j+1}}\pmod{2}$ for every $i,j$. Thus, we conclude that $\angle{A_iOA_{i+1}}\in\mathbb N$ for every $i$.
Now, since $\sum_{i=0}^{18} \angle{A_iOA_{i+1}}=360^\circ$, we must have that $\angle{A_iOA_{i+1}}\equiv 0\pmod{2}$ for every $i$.
Hence, we get $$380^\circ=2(1+2+\ldots +19)=2+4+\ldots + 38\leq \sum_{i=0}^{18} \angle{A_iOA_{i+1}} = 360^\circ,$$which implies a contradiction, meaning there are no such $19$-gon.
b) Answer. $\boxed{\text{Yes.}}$ Consider cyclic $20$-gon $A_0A_1A_2\ldots A_{19}$ with circumcentre $O$.
Let $\angle{A_iOA_{i+1}}=\frac{17}{2}+i$ for every $i=0,1,\ldots,19$, where the indices are taken modulo $20$. Easy to verify that such $20$-gon works.