Let $G$ be a finitely generated group. I wonder if there are examples where:
1) The word problem is known to be solvable in $G$ but there is no algorithm known.
2) The word problem is known to be solvable in $G$ but it is also known that no algorithm for solving the problem can be exhibite.
3) the same as 1) and 2) but with other decisional problems.
Is 2) really possible? The question relies on the difference between "$\exists x$" and "showing an $x$". It seems to me that if $G$ has solvable problem then, by definition, an algorithm $A$ that solves the problem exists. Since algorithms are build up from finite objects, in principle I can enumerate all of them and eventually find $A$ (but how can I be sure that $A$ solves the word problem for $G$?). Am I making a big confusion or the question makes sense?