The use of |a| certainly changes the problem
(i.e. there's only a single answer now),
but I'll have to get back to you later - work duties call.
Separately, I presume you're doing this for a University class, as opposed to just for fun.
My notation has been sloppy with respect to limits.
There are many cases where F(x) can have impulses, such as the rolling a die.
In such cases, there can be a vast distinction between, say, F(a) and F(a-).
This is much like
P(X>x)!= 1-P(X<x), but rather P(X>x)=1-P(X<=x).
Please check with your professor as to your requirements for formality in these, and similar, limit cases.
Added after 4 hours 56 minutes:
When you introduced |a|, you removed the ambiguity as to there being two solutions.
If the restated problem were F[ x | X>(x+|a|) ], then the result would be zero, much like problem 1.
However, you've expressed it just above as
F[ x | X>(x-|a|) ], which does have a region of overlap, depending on |a|.
I think the problem becomes
F[ x| X>(x-|a|) ] = P[ X<x | X>(x-|a|) ] / P[ X>(x-|a|) ] = P[ X<x | X>(x-|a|) ] / (1 - P[ X<=(x-|a|) ]
so, yes ... the answer should be [ F(x) - F(x-|a|) ] / [ 1 - F(x-|a|) ]
As always, all help robustly correct, except where fatally flawed.