Claims about common cycles

Prove or refute the following statements:

Variant
Let \( G \) be a vertex 2-connected graph and \( u, v, w, z \) four of its vertices. Then there is a cycle in \( G \) containing all these vertices.
Solution
This statement indeed does not hold for three vertices. Take a complete bipartite graph \( K_{2{,}3} \) and three vertices from the larger part.

This graph is (vertex) 2-connected, but all cycles are 4-cycles, so none contains all three selected vertices.
Variant
Let \( G \) be a vertex 3-connected graph and \( y, n \) two of its vertices. Then there is a cycle in \( G \) passing through the vertex \( y \) and not passing through the vertex \( n \).
Solution
By removing \( n \), the graph remains two-conneceted. It only suffices to find a cycle passing through \( y \), which is easy. Indeed each pair of vertices lies on a common cycle.
Variant
Each vertex 2-connected graph \( G \) has an orientation of the edges such that for any two vertices \( u \) and \( v \) there exists in \( G \) a unidirectional cycle such that it contains \( u \) and \( v \).
Solution
The statement does not hold. An example is the graph \( K_{2{,}3} \). The vertices of degree 2 in the orientation would have to have one incoming edge and one outgoing edge. Then from these three we can chosen for \( u \) and \( v \) two such that the outgoing edges lead to the same vertex. The cycle specified by \( u \) and \( v \) is then given uniquely and is not be unidirectionally oriented.