Here is an extended hint for proving 2 (an almost complete proof is in the Update below). 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 <a href="http://en.wikipedia.org/wiki/Cyclotomic_polynomial">cyclotomic polynomial</a>, 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 $s$. 

<b> Update</b> Since $2^6=1 \mod 9$, we only need remainders of cyclotomic polynomials modulo $x^3+1$ with coefficients modulo 9. Here are all 28 of them  $$ \begin{array}{l}1,3,{x}^{2},8\,x,1+8\,x,x+1,x+8,4x+5{x}^{2},{x}^{2}+1,\\\1+8{x}^{2}+8x,2+3x+2{x}^{2},2+8x+{x}^{2},2+{x}^{2}+7x,\\\2+2{x
}^{2}+8x,3+2{x}^{2}+7x,\\\4+6x+3{x}^{2},4+7x+4{x}^{2},4+3{x}^{2}+5x,\\\5+2x+7{x}^{2},5+4x+4{x}^{2},5+4{x}^{2}+5x,\\\6+4x+5{x}^{2},6+5{x}^{2}+3x,7+2x+6{x}^{2},\\\ 7+6{x}^{2}+3x,8
+7{x}^{2}+x,{x}^{2}+x+1,{x}^{2}+8x+1\end{array}$$ None of these polynomials become 0 when evaluated at $2 \mod 9$. 

As for the last claim about the size of $q_{mt}(2)$: every root of the cyclotomic polynomial is on the unit circle, so the values at 2 grow with the degree. 

This almost completes the proof of 2 (one needs to show that the set of 28 polynomials is complete, but that can be done by an easy induction).