(a) I can explain the premise behind constructing such a bijection - it is up to you to actually write it up, of course.

First, a little background on power sets. For any set $\displaystyle S$ with a cardinality of $\displaystyle card(S)$, its power set $\displaystyle 2^S$ has the same cardinality as

$\displaystyle \{0,1\}^{card(S)}$

The 0's and 1's are indicator functions telling you whether or not an element of $\displaystyle S$ in a particular subset or not.

For example, consider the set $\displaystyle S=\{x,y,z\}$, which a cardinality of 3 and a power set $\displaystyle 2^S$ which is

$\displaystyle \{$empty set$\displaystyle \}, \{x\}, \{y\}, \{z\}, \{x,y\}, \{y,z\}, \{x,z\}, \{x,y,z\}$

Then, $\displaystyle \{0,1\}^{card(S)}=\{0,1\}^3$

$\displaystyle = \{0,0,0\}, \{1,0,0\}, \{0,1,0\}, \{0,0,1\}, \{1,1,0\}, \{0,1,1\}, \{1,0,1\}, \{1,1,1\}$

Do you see how the set $\displaystyle \{$empty set$\displaystyle \}$ corresponds to $\displaystyle \{0,0,0\}$, $\displaystyle \{1,0,0\}$ to $\displaystyle \{x\}$, so forth?

Extending this principle to infinite sets is possible, so we conclude that $\displaystyle 2^N$ has the same cardinality as $\displaystyle \{0,1\} \times \{0,1\} \times ...$ (multiplied together countable times). But what is this set? This set is all the possible countably sequences of 0's and 1's that exist out there.

Note that if you convert any given real number into binary, you get the same thing: sequences of 0's and 1's. So that's your bijection.