I think the first one is false. If we let $(-1/2, 1/2) \subset \Bbb R$ and $(0,1/4) \subset \Bbb R$, then for $f(x) = x$ defined on $[0,1) \subset M = \Bbb R$, we have
$f^{-1}(-1/2, 1/2) = [0,1/2)$ and $f^{-1}(0,1/4) = (0,1/4)$ both are open in $[0,1)$, but not open in $M$.
I think the second one is right because it is open in the relative sense, and the modification would be (b) applied to (a).
Your counterexample doesn't make sense; the function $f(x) = x$ is certainly continuous at each point of $A \cup B = (-1/2, 1/2)$.
You are, however, correct that statement a) is false, but for a different reason. Consider $M = [0,1]$, and let $A = [0, 1/2)$ and $B = [1/2, 1]$. Then $A \cup B = M$.
What happens if we define $f : A \to \mathbb{R}$ to be constant (say, $f(x) = 0$), and on $B$ define $f(x) = 1$? Are these continuous? Is the resulting function on $A \cup B$ continuous?