The main rule in a triagulation is that each simplex has to be unqiuely determined by its vertices. So for example, you cannot triangulate a circle by taking two one-simplices and identify their endpoints. (This is OK as a cell decomposition, but not as a triangulation.) Hence the simples triangulation of a circle is as a triangle, but of course you can also triangulate it as a square or any $n$-gon. Hence there is no triangulation of $S^1$ with less than 3 one-simplices, but you can have more one-simplices without problems.
The triangulation of the cylinder you consider basically is obtained by taking the simplest triangulation of $S^1$, forming the product with a closed intervall, and then dividing each of the resulting squares into two triangles. It does not matter how you do this subdivision, so the answer to your question (1) is yes. You can do the same thing starting from the triangulation of $S^1$ as a square, thus obtaining a triangulation of the cylinder with $8$ two-simplices and so on.
There always are lots of different triangulations of a topological space, but you can get topological invariants out of them. In this example, the simplest of these would be the alternating sum of the numbers of simplices in each dimension which equals the Euler-characteristic (which happens to be zero for both $S^1$ and the cylinder.