a) No! As a counterexample, consider three small equal circles on the circumference of a sphere. Each of these circles divides the circumference of the sphere into two parts. Call the smaller part for each circle its "inner domain", and assume that the inner domains of these circles are mutually disjoint. Now, if there exists a circle that passes through all three of these circles, it must intersect the sphere in at least three points (it must intersect each inner domain at one point) which is impossible, since any circle that is not on the circumference of a sphere meets it in at most two points.

b) Yes. Consider three circles $C_R$, $C_B$ and $C_G$ in the space such that no two of them are co-planar. Denote the plane containing $C_i$ by $\pi_i$ (for $i \in \{R, B, G\}$). For each circle $C_i$, let $X_i$ be a point in the plane $\pi_i$ lying inside $C_i$, and choose them so that these three points are not collinear, so they form a plane, say $\pi$. Due to our assumption plane $\pi$ meets each circle at exactly two points. Denote these intersection points by $\{G_1, G_2\}$, $\{B_1, B_2\}$ and $\{R_1, R_2\}$, respectively. Now, the goal is to find a square in $\pi$ that intersects each of the three segments $G_1G_2$, $B_1B_2$ and $R_1R_2$ at exactly one inner point (not an endpoint). Obviously, since this square separates each pair of points, it must pass through all three circles and hence is the desired square. To show its existence, first a lemma is proved.

Lemma. For points $X$, $Y$, $A$ and $B$ in the plane where $XY$ is not parallel to $AB$, a square exists that passes through points $X$ and $Y$ and separates $A$ from $B$.

Proof. First, suppose that $XY$ is not perpendicular to $AB$. If line $XY$ separates $A$ from $B$, a large square with $X$ and $Y$ both on one of its sides works.

On the other hand, if both points $A$ and $B$ lie on one side of line $XY$, depending on the distance from $A$ and $B$ to $XY$ and the position of the feet of perpendicular lines from these two points to line $XY$, a suitable square as depicted in the following figures can be found.

If $XY \perp AB$, the argument must be changed only when the intersection point of $AB$ and $XY$ lies outside of both segments. In this case, a suitable square can be found like the following figure.

Since points $R_1, R_2, B_1$ and $B_2$ are not collinear, points $X$ on $R_1R_2$ and $Y$ on $B_1B_2$ exist such that line $XY$ does not contain any point from $R_1R_2$ and $B_1B_2$ other than $X$ and $Y$ and also, $G_1G_2$ is not parallel to $XY$. According to the lemma, a square passing through $X$ and $Y$ (and no other point on $R_1R_2$ and $B_1B_2$) and separating $G_1$ from $G_2$ exist, which completes the proof.