For 1 and 3, yes. (For 3 assuming both are positive, technically one positive would suffice but then it is essentially reduced to 1, while of course both negative will not work.)

And, note that this is only interisting for natural coefficients, with integers you can write any integer whatsoever, and this is very classical, not even sure how to call this, a direct consequence of Bezout's identity maybe, or an immediate consequence of results on linear diophantine equations.

For 2. The main keyword (with natural coefficients) here is Frobenius problem, but mainly for more than two numbers. Also Coin Problem, Postage Stamp Problem [though it seems some use this to refer to a realted but distinct problem where there is a restrition on the sum of coefficients, on the grounds that an enveloppe can only hold a given number of stamps] or Chicken McNuggets Problem are common more playful names. See http://en.wikipedia.org/wiki/Coin_problem

The solution for 3 is due to Sylvester.

If you have more than two coprime natural numbers $a_1,\dots , a_k$ for integer coefficients you still can write everything. But for natural number coefficients you can only write everythong starting from a certain number $F(a_1,\dots , a_k)+1$ on. To determine the optimal value here is *hard* (even algorithmically).

This is the Frobenius Problem, and $F(a_1,\dots , a_k)$, the largest number that cannot be written in this form, is called the Frobenius number of $a_1, \dots, a_k$ (or also of the numerical semigroup generated by $a_1, \dots, a_k$ that is just the additive semigroup generated by the $a_i$)

It was in a certain sense solved recently for three numbers by Fel .

There are numerous other contributions on this problem and a lot of literature on numerical semigroups (that appear naturally in other areas of math, see the link above for some info), in particular there is a rather recent (2005) monograph on this subject by J. Ramírez Alfonsín "The Diophantine Frobenius problem."