I think I've found a proof to 1. (It's quite similar to Jean's comment above.)
Assume $G$ is the smallest (finite non-empty) graph in which every vertex has degree at least 3 and there are no chords.
Observation: Every edge in $G$ belongs to at most one cycle.
Let $C$ be a cycle in $G$ (if no cycles exist then we violate the degree condition). We identify the vertices in $C$ in $G$, thereby creating a new graph $H$ with strictly fewer vertices.
The vertex formed by the identification, $x$ say, is a cut-point of $H$, by the above observation. Thus, if $H$ contains a cycle with a chord, then so does $G$. Thus $H$ does not contain a cycle with a chord. We also observe that $x$ has degree $\geq 3$, since they neighbours of the vertices of $C$ in $G$ must all be distinct.
Hence $H$ is a graph with minimum degree $3$ and no cycles with chords, giving a contradiction. We conclude that there is no minimum counter-example.
Note, however, that there are infinite trees without chords (in fact, without cycles) and minimum degree $N$ for any $N \geq 3$.