The answer to the second question is no.

It is known that if f has IVT but it is not continuous at a point it does not have a limit at that point.

If the condition "l(x) being in R-{f(x)} for all x in R" means that the limit of f at x is not equal to the value of f at x, then f has a removable discontinuity at every point of R; thus, the set of removable discontinuities of f is uncountable. However, it is known that the set of removable discontinuities of every real function is countable (https://en.wikipedia.org/wiki/Classification_of_discontinuities#Counting_discontinuities_of_a_function) . Therefore, such a function does not exist.

For the second question the response is clear! For the first, my response "no" was justified by the Blumberg Theorem: Every real function f:R->R has a continuous restriction f#:D->R, where D is a dense subset of R. I don't remember exactly the statement but I think D must be also countable.

I knew only for monotonic functions from R to R the set of discontinuities is at most countable. I didn't consider important that here the discontinuities were removable !

Regarding the cardinality of the set D in Dinu's answer, this paper https://math.wvu.edu/~kciesiel/prepF/139BlumbergThm.pdf explains (see Theorem 2) why D must be countable.

Also, here https://math.stackexchange.com/questions/1275122/can-we-construct-a-function-that-has-uncountable-many-jump-discontinuities?rq=1 (first answer), it is proved that the set of jump discontinuities of a real function, not necessarily monotonic, is always countable. The proof is similar if the discontinuties are removable (see the note after the first answer).

Hayjaa Khudhair Dakhil , your post is not readable, but it seems that you claim that the function f(x) = x+1 if x =/ 0 and f(0)=0 satisfies the condition of the question. Your function is continuous everywhere except 0; it does not satisfy the condition of the question.

Every continuous function has the IVP, but not every function with the IVP is continuous. For example, the function f(x)=sin(1/x)

is not continuous on x=0.