(b) $f(x)=1$ is periodic, but not cool, since for all $b$, $f(x+b)=f(x)$, and $f(x)$ is not odd.

(a) By definition, then there exists $a,b$ such that \[ f(x+a) = f(-x + a) \ \text{and} \ f(x + b) + f(-x + b) = 0 \]for all $x \in \mathbb{R}$. Now, check that \begin{align*} f(x + b + a) &= -f(-(x + a) + b) \\ &= -f(-(x -b +2a) + a) \\ &= -f(x - b + 2a + a) \\ &= -f(x + 3a - b) \\ &= -f((x + 3a - 2b) + b) \\ &= f(-(x + 3a - 2b) + b) \\ &= f(-x - 3a + 3b) \\ &= f(-(x - 3b + 4a) + a) \\ &= f(x - 3b + 4a + a) \\ &= f(x - 3b + 5a) \end{align*}for all $x \in \mathbb{R}$.

$f(2a-x)=f(x)$ and $f(2b-x) = -f(x)$. Hence $f(2b+x-2a) = -f(2a-x) = - f(x)$ Let $c = 2b-2a$. So $f(c+x) = -f(x)$ and hence $f(2c+x) = -f(c+x) = f(x)$ and hence $4(b-a)$ is a period

To complete these proofs, note that if $a=b$ then $f(x+a)$ is odd and even, hence is allzero, so $f(x)=0$ which is periodic.