> Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)?

floor(x/2) = 0 on the interval [0, 2), so the expression reduces to x/2.

> And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))?

If y = x/2 - floor(x/2) is periodic, then e^y = e^(x/2 - floor(x/2)) must be periodic as well, with the same period.