Intriguingly, the last one is the trickiest. And while you have used the essential trick, you didn't do it completely right.
b) should be rather trivial, just let $g(x)=d-\frac{d-c}{b-a}\cdot(x-a)$.
For a) one can take something like $f(x)=\tan (\frac \pi2 x)$ or alternatively,
$$f(x)=\begin{cases}0&\text{if }x=0\\\frac1x-1&\text{otherwise}\end{cases}$$
The trick with c) i sthat we want to get rid of the one endpoint $1$, so we might want to leave most points fixed and move $1$ to $\frac12$. Howevre, this implies we cannot leave $\frac12$ fixed and have to move it aways, say to $\frac13$. And so on. Explicitly
$$h(x)=\begin{cases}\frac1{n+1}&\text{if } x=\frac1n,n\in\mathbb N\\
x&\text{otherwise}\end{cases}$$
I recommend you lookup "Hilbert's hotel".
In all cases above you should take a moment to check the necessary properties. That is, to show that $\phi\colon A\to B$ is a bijection, you need to show that
- for each $x\in X$, the value $\phi(x)$ is indeed $\in Y$
- $\phi$ is onto, that is for each $y\in Y$ we can exhibit some $x$ with $\phi(x)=y$
- $\phi$ is 1-to-1, that is if $\phi(x_1)=\phi(x_2)$ for $x_1,x_2\in X$ then $x_1=x_2$.
