Prove $f$ is periodic if $ \int_{a}^b f(x)dx = 2 $ and $ \int_{y}^z f(x)dx = 1 $ $z,y\in (a,b)$ and $z-y=(a-b)/2$ if $ \int_{a}^b f(x)dx = 2 $ and for every $y,z \in (a,b)$ ($y$ smaller) such that $z-y = (a-b)/2$
we have  $ \int_{y}^z f(x)dx = 1 $ how do we prove f is periodical and find the period?

Let's assume $a<b$. By "periodic(al)" I assume you mean that: 
$$f(x) = f(x+(b-a)/2) \quad \forall x \in [a, a + (b-a)/2]$$
Case 1:
If $f$ is not necessarily continuous, then the result is not true.  That is because you can take any $f$ that works, then change its value on a finite number of points so that the new function is not periodic.  But the integrals are unchanged. 
Case 2:
If $f$ is continuous, you can define $H(y)=\int_y^{y+(b-a)/2}f(x)dx$ and then see what happens when you use the Fundamental Theorem of Calculus.
