As I alluded to in a comment to one of your previous questions, many people have thought about how to explicitly compute fundamental domains, homology, etc., for congruence subgroups of $SL_2(\mathbb Z)$: this leads to the theory of modular symbols, which is not only a theory, but serves as the foundation for all computational work on modular forms too.
(See papers of Mazur and Manin from the 1970s, and for the computational aspects, more recent writings of Cremona, William Stein, and others.)
On a related note, let me remark that
it it actually quite feasible, and reasonable, to generate a fundamental domain for a finite index subgroup by translating the fundamental domain for $SL_2(\mathbb Z)$.
For a carefully worked example, see here (especially the discussion on page 6).
For (1) the answer is "yes". Since the surface has finitely generated fundamental group, there is a finite collection of disjoint embedded bi-infinite geodesics that cut the surface into a collection of ideal (or hyperideal) polygons, each with at most one cone point in its interior. (There is a special case when the cone point has angle $\pi$, which I leave as an exercise.) The statement now follows from the Seifert-van Kampen theorem.
For (2) I read your question as "can a Dirichlet domain for a surface (as above) have material vertices?" and the answer is "yes, it may". To see this, consider the surface made by doubling an ideal triangle across its boundary. If we take the origin to be the centre of one of the triangles, then the Dirichlet domain has three ideal vertices and three material vertices.
For (3) the answer is "not in general". To see this, consider the surface of (2) where we "open" the cusps, replacing them by (identical) funnels. Again the surface has a three-fold symmetry about the centres of the (now hyperideal) triangles. The Dirichlet domain has three material vertices, six ideal vertices, six material edges, and three ideal edges.
The property you want - disjoint material edges glued in pairs - feels very similar to "Schottky, with all circles perpendicular to a single circle". Even this only gets you some Dirichlet domain with the desired form, not all. I am not sure if this property has a name, but I would look at Marden's book, or Maskit's, as a first reference.
For (3) "updated" the answer is "not in general". That is, there is a hyperbolic surface $S$ with infinite volume, with finitely generated (and free) fundamental group, so that all Dirichlet domains have material vertices.
Consider the surface $T$ given by (3) (first version) above. Let $x$ in $T$ be one of the points with three-fold symmetry. Let $D \subset T$ be a very small round disk about $x$. Let $S$ be the surface obtained by doubling $T - D$ across $\partial D$. Uniformise $S$. Let $\gamma$ be the image of $\partial D$ in $S$. Since $\gamma$ is fixed by a reflection in $R$, it is also a hyperbolic geodesic. Since the radius of $D$ was very small, the geodesic $\gamma$ is very short.
I claim that $S$ has the desired property - that is, all Dirichlet domains have material vertices. The proof is harder than that for (3).
Best Answer
Yes, this is true. Topologically, one may find a locally finite collection of properly embedded arcs in a connected surface whose complement is homeomorphic to $R^2$. Then make each of these arcs geodesic in the hyperbolic metric. The complement will be the fundamental domain of the type you want.
Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a classification of connected surfaces. In Theorem 3 of that paper, he explains how to construct all surfaces. A planar surface $\Sigma\cong S^2-X$ is obtained by removing a totally disconnected compact set $X\subset S^2$ from $S^2$. As explained in Prop. 5 of the paper, one may consider the totally disconnected set $X\subset S^2$ to be a subset of the Cantor set, and therefore a subset of the interval (including the endpoints) $X\subset I\subset S^2$. Then the properly embedded arcs $I\cap (S^2-X)$ give a decomposition of $\Sigma$ into $R^2$.
If the surface is non-planar, then one removes from $S^2-X$ a properly embedded countable collection of disks $D_1,D_2,\ldots$, and makes identifications of their boundaries. We may assume after an isotopy that these disks are all centered on $I$, and that the identifications either identifies antipodal points, or identifies two disks which are adjacent along a component of $I-X$ with a $\pi$ twist. The complement $U=S^2-(I\cup_i D_i)$ is again homemorphic to $R^2$. If we identify antipodal points of $D_j$, then this identifies two arcs in the boundary of $U$ to obtain an open Mobius strip. We add two arcs connecting antipodal points of $D_j$ to a point $x\in X$ at the end of the interval of $I-X$ which intersects $D_j$, which forms a single arc after identification of antipodal points of $D_j$, and cuts the Mobius strip back up into $R^2$. If adjacent disks $D_i, D_{i+1}$ are identified, then the complement $U$ gives a punctured torus. We add 4 arcs connecting these points to $x$ (again, $x\in X$ is at the end of the interval of $I-X$ containing $D_i$), cutting the surface into $R^2$ again. Continuing in this fashion inductively, we get a locally finite collection of arcs cutting the surface up into $R^2$.