Let $G$ be a simple cubic graph which has a Hamiltonian circuit $C$.  In general, it is not possible to find a second Hamiltonian circuit which contains all the chords of $C$.  For example, the Wagner graph with $C$ being the outer circuit in the picture contains no such second Hamiltonian circuit.  

[![Wagner graph][1]][1]

But it is possible to add an additional chord so that the resulting graph contains a second Hamiltonian circuit including all chords, as in the second picture?

[![Wagner with chord][2]][2]

Our first question is:

>**Question 1 (EGME)** Let $G$ be a cubic graph which has a Hamiltonian circuit $C$.  Is it always possible to add a chord of $C$ in $G$ so that the resulting graph has a Hamiltonian circuit containing all the chords? Please provide proof or counterexample.

Our second question is

>**Question 2 (EGME)** If we add ANY new set of chords of $C$ in $G$ (as in question 1) which are a perfect matching in the resulting graph (which should be simple), is there a second Hamiltonian circuit which contains all of the chords in the original set? Or equivalently, let $G$ be a 4-regular Hamiltonian graph $G$ with a Hamiltonian circuit $C$, and such that the chords of $C$ can be partitioned into two perfect matchings $P$, $Q$.  Is there a Hamiltonian circuit of $G$ which contains all the edges in $P$ ($Q$).  If not, under what conditions is this true?

Computer experiments seem to suggest this might be the case.

  [1]: https://i.sstatic.net/wxHgu.png
  [2]: https://i.sstatic.net/TimvH.png