Official solution
Here both R1 and R2 are symmetric as for any (x, y) \[ \in \] R1, we have (y, x) \[ \in \] R1 and similarly for any (x, y) \[ \in \] R2, we have (y, x) \[ \in \] R2
In R1, (b, c) \[ \in \] R1, (c, a) \[ \in \] R1 but (b,a) \[ \notin \] R1
Similarly in R2, (b, a) \[ \in \] R2, (a, c) \[ \in \] R2 but (b, c) \[ \notin \] R2
\[ \therefore \] R1 and R2 are not transitive.