You might have noticed that the difference between your two sequences is $0,1,2,3,4,5,6,7,8,\ldots$ and that the optimal configurations seem to be the same for both problems.

This is true in general, and is a nice application of Euler's formula $V-E+F=2$. (NB this requires showing that in an optimal configuration the edges form a connected component.) Here $V=n$, the number of edges is $E$, and the number of triangles, call it $t$, is $F-1$ because we must count the exterior of the graph as a face. So the difference between the edge and triangle maxima is $(n-1)$ — at least assuming we can prove it's never to our advantage in either problem to have holes bigger than a unit triangle in the picture, which seems clear but may be annoying to prove rigorously. [**EDIT** see below on this point.]

Another standard graph-theory formula that applies here: the sum of all the faces' edge-counts is $2E$. [Proof: count in two ways the pairs $(e,f)$ where $f$ is a face and $e$ is one of its edges.] In our setting all but one of the faces has $3$ edges, so $2E=3t+p$ where $p$ (for perimeter) is the number of outside edges (counting an edge twice if both sides abut the infinite face, as happens for $n=2$, and again assuming no internal holes). So $t+2 = 2n-p$, and the problem comes down to minimizing $t$. It certainly looks plausible that the "penny spiral" does this, but it's getting late so I'll leave it as an exercise :-) This would identify your first sequence with OEIS A047932, and thus determine the second sequence as well.

**UPDATE** Denote the sequences in question by $s_1(n)$ and $s_2(n)$ respectively. As pointed out in G.Zaimi's accepted answer, the formula $s_1(n) = \lfloor 3n-\sqrt{12n-3}\rfloor$, consistent with the original proposer's guess/question, follows from a much more general result of Harbroth that this is the maximal number of times that the minimum distance can occur in *any* configuration of $n$ points in the plane. I looked up the solution (which is freely available online), and it seems to use ultimately the same technique: see the reference to the "Eulerschen Polyedersatz" before equation (3). Using the Euler "Polyedersatz" we also deduce $$s_2(n) = s_1(n) - (n-1) = \lfloor 2n-\sqrt{12n-3}+1\rfloor,$$ answering the second question. Indeed Euler says that $s_2(n) \leq s_1(n) - (n-1)$ with equality iff there's an optimal configuration without holes bigger than a unit triangle; and Harbroth's solution gives such a configuration.