For the first question, we can alternatively directly count both E and O like so:
Generally, the number of strings in $\{0,1\}^{17}$ with $k$ $1$'s is $\displaystyle\binom {17}k $, as we're ranging over all possible spots to put the $k$ ones in. So E equals the sum of this over all even numbers, and similarly O is the sum over all odd numbers. We can calculate
E - O $ = \displaystyle \sum_{i=0}^{8} \displaystyle \binom {17}{2i} - \sum_{i=0}^{8} \displaystyle \binom {17}{2i + 1} = \sum_{i=0}^{17} \displaystyle (-1)^i \binom {17}{i} = \sum_{i=0}^{17}\displaystyle 1^{n-i}(-1)^i \binom {17}{i} = (1-1)^{17} = 0^{17} = 0.$
However, this approach doesn't seem to lend itself to the next question directly, unless you notice that a sum is even iff the the number of $1$'s plus the number of $3$'s is even. In this case, you can consider $1$ and $3$ as one element, and $0$ and $2$ as one element (sorting them by parity). We're essentially looking for E here and so we can apply our result that E = O from above. Combined with the knowledge that E $+$ O = our total number of strings, $4^{17}$, we see that E = $\dfrac{4^{17}}{2}$.
A computational approach is helpful and possible, but the often superior option is to come up with what is called a combinatorial proof based on direct reasoning, as in Arturo or lhf's answers. As Stanley writes in enumerative combinatics,
Not only is the above combinatorial proof much shorter than our previous proof, but also it makes the reason for the simple answer completely transparent. It is often the case, as occurred here, that the first proof to come to mind turns out to be laborious and inelegant, but that the final answer suggests a simple combinatorial proof. (page 12)