Answer
a) Yes.
b) No.
Work Step by Step
The points are on the graph if and only if $x^2+y^2=1.$
a) $x^2+y^2=(\frac{1}{2})^2+(\frac{1}{2})^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\ne1.$ Hence it is not on the graph.
b) $x^2+y^2=(\frac{1}{2})^2+(\frac{\sqrt3}{2})^2=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1.$ Hence it is on the graph.
