Reformulating your second question concerning the distance of the fraction $f(n)={3^n-0.5\over2^n-0.5}$ to the next integer in terms of $g(n)=(3/2)^n$ shows first that
$$ f(n) = g(n) + \frac12 (3/4)^n - \varepsilon_n
$$
where $\varepsilon_n < \frac12(3/4)^n $
Then [including also the floor-expression into an inequality] we can write
$$ \left[ \lfloor g(n) \rfloor < \right] \qquad \qquad g(n) < f(n)<g(n)+(3/4)^n$$
The Waring-conjecture says now
$$g(n)+(3/4)^n < \lceil g(n) \rceil $$
Assuming the truth of the Waring-conjecture, your function $f(x)$ is then smaller than the next integer because $g(n)< f(n) < g(n) + (3/4)^n$ .
Unfortunately the Waring-conjecture is not yet proven, and the best proven bound known so far is that $g(n)+0.577^n \lt \lceil g(n) \rceil $ so your second question belong to a "still open problem".
The first question can be related to the problem whether with the function $h(n,m)={m3^n-1\over m2^n-1} $ we have for all $n>1$,$m \ge 1$ with some suitable integer $b$ always $$ 2^{k-1} \lt h(n,m) \lt 2^k $$
Here we have a proof of Ray Steiner (1977) where he was looking for cases where $h(n,m)=2^k$ for making a cycle in the Collatz-problem possible, that this equality never occurs for $n>2$,$m>1$, cited and paraphrased in some literature about the collatz-problem, for instance by
John Simons (2000,2002). Your question would ask for $h(n,m)$ with $m=2$ and this case is included in the Steiner-proof.
Update: Since in your comment you mention that this question stems from studies of the Collatz-problem, I've tried to put things together as far as I've understood them and provide a proof for the noexistence of 1-cycles which is intended to be more readable than even the latest version of J. Simons' article. See at
my homepage
Update 2:
To solve the first question we assume that with $S=k+N$ we have a value in $k$ such that $2^S>3^N$ .
Then we rewrite
$$ \large \begin{array}{rll} 2^k &\overset?>& { 3^N-1/2\over 2^N-1/2} \\
{ 2^S \over 3^N} &\overset?>& {1 -1/2/3^N \over 1-1/2/2^N} \\
\end{array}$$ and logarithmize
$$\begin{array}{rll}
S \log 2 - N \log 3 &\overset?>& \log(1 -1/2/3^N ) - \log(1-1/2/2^N) \\
\end{array} \tag 1$$
By G. Rhin (cited by J.Simons) we know that
$$ S \log 2 - N \log 3 > { 1 \over 457 \cdot N^{13.3} } \tag 2$$
Now we compare the rhs of (2) with the rhs of (1). Of course, if rhs of (1) is smaller than rhs of (2) then the inequality (1) is also true.
$$ { 1 \over 457 \cdot N^{13.3} } \overset?>\log(1 -1/2/3^N ) - \log(1-1/2/2^N)
$$
The logarithms in the rhs expand to series with quickly vanishing terms of order of geometric decrease:
$$ { 1 \over 457 \cdot N^{13.3} } \overset?>
(1/2/2^N) - (1/2/3^N) + (1/2/2^N)^2/2 - (1/2/3^N)^2/2 + ... + \varepsilon \tag 3 \\
$$
and we see, that in the rhs the size diminuishes exponentially with $N$ while on the lhs only polynomially in $N$ (to power of $13.3$) Thus there shall be one crossover-point in $N$ after which the rhs shall always be smaller than the rhs.
If we simply test numerically small $N$ from $1$ to $100$ we find for $N=96$ by comparision $$N=96: \text{lhs=} 9.45967902141 E-30 > 6.31088724177 E-30 \text{=rhs}$$ and we can complete the inequality
$$ S \log 2 - N \log 2 > \frac 1{457} \frac 1{N^{13.3}} > \log(1-1/2/2^N) - \log(1-1/2/3^N) \qquad \text{for }N \ge 96\tag 4 $$
This implies for $N \ge 96$
$$
\implies {2^S \over 3^N } > {1-1/2/3^N \over 1-1/2/2^N } \qquad \qquad \text{for }N \ge 96$$ and
$$
\implies 2^k > {3^N-1/2 \over 2^N-1/2 } \qquad \qquad \text{for }N \ge 96 \tag 5
$$
Checking the remaining $94$ choices for $N<96$ directly completes then the proof for the full range of $N$. One should not forget to mention: this is based on the result of G. Rhin in 1984 (see the reference in the article of J. Simons).
