Why there are no approximation algorithms for SAT and other decision problems?
Approximation algorithms are only for optimization problems, not for decision problems.

Why don't we define the approximation ratio to be the fraction of mistakes an algorithm makes, when trying to solve some decision problem?  Because "the approximation ratio" is a term with a well-defined, standard meaning, one that means something else, and it would be confusing to use the same term for two different things.

OK, could we define some other ratio (let's call it something else -- e.g., "the det-ratio") that quantifies the number of mistakes an algorithm makes, for some decision problem?  Well, it's not clear how to do that.  What would be the denominator for that fraction?  Or, to put it another way: there are going to be an infinite number of problem instances, and for some of them the algorithm will give the right answer and others it will give the wrong answer, so you end up with a ratio that is "something divided by infinity", and that ends up being meaningless or not defined.

Alternatively, we could define $r_n$ to be the fraction of mistakes the algorithm mistakes, on problem instances of size $n$.  Then, we could compute the limit of $r_n$ as $n \to \infty$, if such a limit exists.  This would be well-defined (if the limit exists).  However, in most cases, this might not be terribly useful.  In particular, it implicitly assumes a uniform distribution on problem instances.  However, in the real world, the actual distribution on problem instances may not be uniform -- it is often very far from uniform.  Consequently, the number you get in this way is often not as useful as you might hope: it often gives a misleading impression of how good the algorithm is.

To learn more about how people deal with intractability (NP-hardness), take a look at Dealing with intractability: NP-complete problems.