real analysis - Let $f: [0,1] rightarrow mathbb{R} $ be continuous with $f(0) = f(1)$ *note, there is a part b*

(a) Show that there must exist $x,y \in [0,1] $ satisfying $|x-y| = \frac{1} {2}$ and $f(x) = f(y)$

I can start by defining a function $g(x) = f(x + \frac{1} {2}) - f(x)$ to guarantee an $x,y$ so that $|x-y| = \frac{1} {2}$ But how do I show that $f(x) = f(y)$?

(b) Show that for each $n \in \mathbb{N}$ $\exists x_n ,y_n \in [0,1]$ with $|x_n - y_n| = \frac{1} {n}$, and $f(x_n) = f(y_n)$

Actually I'm not sure where to start here. Any help is greatly appreciated.
Thanks!