Determining covalent bond strengths without hybridisation
As I already stated, asking this question is wrong on so many levels, but let's just have a look anyway.

We are assuming a set of $3$ rigid $\ce{2p}$ and $1$ rigid $\ce{2s}$ orbitals at carbon. With rigid I mean, hybridisationnote is forbidden - which is wrong, $\text{[very]}_n$ wrong.

As a consequence, a head on overlap between the hydrogen $\ce{s}$ and the carbon $\ce{p}$ results in three bonds, each perpendicular , i.e. $90^\circ$, against each other. This is the feature of the degenerate $\ce{p}$ orbitals. (Answer 1 would be correct.)

As a conclusion from that, the molecule cannot adapt a tetrahedral geometry, because with this geometric arrangement, it is impossible to form a $S_4$ symmetry axis, which is the main feature of the $T_\ce{d}$ or $T$ point groups. Answer 3 would be wrong.

The bond between the carbon $\ce{s}$ and hydrogen $\ce{s}$ also needs to be weaker than the others, since these orbitals are totally symmetric. That means there is a good portion of the carbon $\ce{s}$ orbital already blocked by the other hydrogens. On the other hand, the orbital overlap of the carbon $\ce{p}$ and the hydrogen $\ce{s}$ is quite optimal, since the main contribution of $\ce{p}$ orbitals lie outside of the carbon $\ce{s}$ orbital. (Answer 2 would be correct.)

This $\ce{s-s}$ bond would also be quite wobbly, because neither of the orbitals is directed. Therefore the position of the last hydrogen can easily change without effecting the orbital overlap, and in first approximation the bond strength. (Answer 4 would be correct.)

As a last comment, the molecule in this conformation is not stable, probably not even as an highly excited state. The kinds of thought games actually ruin the more important thing of understanding how orbitals behave. Orbitals are not rigid. They are quite flexible in the end.



Note:
Hybridisation in the generic sense means the linear combination of atomic orbitals of one atom, that contribute to a molecular orbital.