I think both (a) and (b) are true, because all the sets in the questions are infinite uncountable, so we can construct a bijection map. For question a, [0,1] and [0,1] are exactly the same set so I am not sure how the map [0,1] to [0,1] union A. If I construct a function K defined as follow, K(x) = {f(x)=(x-1)/2 if x belongs to A, g(x)=x/2 for all x belongs to the image of f(x), and h(x) if otherwise. Is it a bijection map? For question (b), I am not sure how to show there exists a bijection map.
$\mathfrak{c} = |[0,1]| \le|[0,1\ \cup A| \le \mathfrak{c} + |A| \le \mathfrak{c} + \mathfrak{c}=\mathfrak{c}$ so yes for a).
$\mathfrak{c}=|[x_0-\frac{r}{2}, x_0 + \frac{r}{2}]| \le |B| \le |[0,1] = \mathfrak{c}$ so yes for b) too.