Here is an extended hint for proving 2. If $3^s$ base 2 is periodic, then you can represent it as $u(1+2^m+2^{2m}+...+2^{(t-1)m})=u\frac{2^{tm}-1}{2^m-1}$ for some numbers $u,m,t$. Therefore if $q_n(x)$ denotes the $n$-th cyclotomic polynomial, then $q_{tm}(2)$ must be a power of 3 (as a divisor of $3^s$). But it looks like $q_n(2)$ is never 0 modulo 9 (this should be possible to prove rigorously but I do not have time). Hence $q_{tm}(2)$ must be equal to 3 or 1 which gives a bound on $s$.
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