So I have finitely presented group with 2 generators. Can I solve word problem in it (check if two words X and Y are actually the same element of my group)?
It is a classical result, originally by Higman-Neumann-Neumann, that every finitely presented group embeds in a 2-generator finitely presented group. In particular, every finitely presented group with unsolvable word problem embeds in a 2-generator group. Given that having solvable word problem is preserved by subgroups, such 2-generator groups must also have unsolvable word problem.
Moreover, this construction is uniform: given a finite presentation P, we can construct a new 2-generator finite presentation Q into which P embeds (as well as an explicit embedding). Also, Q has m+n relators, where m is the number of generators of P, and n is the number of relators of P. However, the relators of Q can get very long.
You can find their original paper at: G. Higman, B. H. Neumann, H. Neumann, "Embedding theorems for groups", J. London Math. Soc. 24, 247-254 (1949).
In general the answer is no.
A counterexample can be found in the paper by W. Boone "The word problem" (Ann. of Math. (2) 70 (1959) 207–265).
Quoting from page 210:
"It thus follows from Result c (using the embedding result of  noted above) that one can exhibit a finite presentation of a group consisting of two generators and thirty-two defining relations and having an unsolvable word problem."