For Q4, try to directly solve x+c = f(x). You end up with a quadratic involving c and a. Consider the discriminant D of that quadratic.
If there's to be an intersection regardless of c, then we must have D>=0 for all c. But D is a quadratic function of c, so for that to be true, the discriminant of D (considered as a function of c) must be >=0. This gives a quadratic inequality in a, which eventually gives the desired result.
For Q6, try a substitution x=y-a for a suitable choice of a.