I assume "dyadic" here means "binary". It is easy to prove by induction on the binary length of numbers that \[ a(x+y)\leq a(x)+a(y).\] Let $ n=2^{b_1} + 2^{b_2} + \dots + 2^{b_m}$ be binary representation of an integer $ n$, where $ b_1<b_2<\dots<b_m$. Then $ a(n)=m$ and \[ n^2 = \sum_{i=1}^m 2^{2b_i} + \sum_{i=1}^m\sum_{j=i+1}^m 2^{b_i+b_j+1},\] implying that a proof for the problem a.: \[ a(n^2)\leq \sum_{i=1}^m 1 + \sum_{i=1}^m\sum_{j=i+1}^m 1=\frac{m(m+1)}{2}.\] To establish b., it is enough to show that there is infinitely many sets of numbers $ b_1<\dots<b_m$ for which $ |\{ 2b_i, b_i+b_j+1 \}|=\frac{m(m+1)}{2}$. It is clear that this requirement holds for any sequence $ b_1<\dots<b_m$ that grows fast enough. For example, any sequence for which $ b_{i+1}>2b_i$ for every $ i=1,\dots,m-1$ works. To prove c., it is enough to take \[ n_i = 2^{2^i-1} - \sum_{j=1}^i 2^{2^i-2^j}.\]