Your post is a little jumbled, but let's look at two things and I think you will understand which option in your post is the correct answer.
First of all, look at the definition of the Mean Value Theorem:
If a function f is differentiable on an interval (a, b), and it is also continuous on [a, b], then there is at least one point c where the slope at c is equal to the secant line between a and b.
Writing that mathematically:
f'(c) = [f(b) - f(a)]/(b-a)
the slope at c = the slope of the secant line between a and b
In this problem, the function f(x) = (x-3)-2, and hopefully you recognize that this function is not continuous in the interval [2, 5]. Specifically, it's discontinuous at x=3. <+++++ PART 1 of the answer
Secondly, if you calculate the slope of the secant line between 2 and 5, you'll find that it's -0.25.
If you try to find a point c on this function where the slope is equal to -0.25, you'll find that it is outside the interval [2, 5]... in other words, there's no point c in the interval [2, 5] where the Mean Value Theorem holds true. But, the Mean Value Theorem is not contradicted, because the Mean Value Theorem only applies to functions that are continuous on the interval <++++++ PART 2 of the answer
So hopefully now you can see that "This does not contradict the Mean Value Theorem since f is not continuous at x = 3"
This means, the result from Part 2 doesn't contradict the Mean Value Theorem, because the Mean Value Theorem only applies to continuous functions.
Hope that helps, and feel free to contact me if you have any more questions.