The first relation $b(0)=0$ is trivial.

By the condition, $a(n)$ is the product of number of consecutive zeroes plus $1$, if we write $n$ in binary. For instance, if $n=1220$, we have $n=100 1100 0100$ in binary, thus $a(n)=3\times 4\times 3=36$.

We need a lemma: Lucas' Theorem: $\binom{a}{b}$ is odd if and only if $a\&b=b$ where $\&$ is bitwise AND operation. Let $S_n$ be the set of $i$ if $i\&n=i$.

Now the second relation is clear. Notice that if $k\in S_{2n+1}$ if there exist $k'\in S_n$ such that $k=2k'$ or $k=2k'+1$. The even ones consists of exactly the elements in $S_{2n}$. So we can divide $S_{2n+1}$ into odd part and even part, We have

$$b(2n+1)=\sum_{k\in S_{2n+1}}a(k)=\sum_{k\in S_{2n+1},k\equiv 0\mod 2}a(k)+\sum_{k\in S_{2n+1},k\equiv 1\mod 2}a(k)$$
We know that
$$\sum_{k\in S_{2n+1},k\equiv 0\mod 2}a(k)=\sum_{k\in S_{2n}}a(k)=b(2n)$$

And since the odd numbers, when deleting the ending $1$ the $a(\cdot)$ value won't change. Thus, 
$$\sum_{k\in S_{2n+1},k\equiv 1\mod 2}a(k)=\sum_{\in S_{2n}}a(k)=b(n)$$
Therefore we have $b(2n+1)=b(2n)+b(n)$

For the third, I am still figuring it out.