For the second question, if $b = \sqrt{a}$ is an irrational number with very good rational approximations, there won't be an upper bound. For example, if
$$ \dfrac{p}{q} + \dfrac{1}{q^r} > b > \dfrac{p}{q}$$
for coprime positive integers $p,q$ and integer $r>1$, then taking $n=q^r$ we have $\lfloor n b \rfloor = p q^{r-1}$ and
$\gcd(\lfloor n b \rfloor, n) = q^{r-1} = n^{1-1/r}$. For example, you could take
$$b = \sum_{j=0}^\infty 2^{-(r+1)^j}$$
and have this with $q = 2^{(r+1)^m}$ for all positive integers $m$.

Of course, Roth's theorem says for an algebraic number $b$ there will be only finitely many such $q$.

EDIT: Conversely, suppose $\gcd(\lfloor nb \rfloor, n) = g$, with $n=gu$ and $\lfloor nb \rfloor = g v$ for coprime positive integers $u$,$v$. Thus
$$ \dfrac{v}{u} + \dfrac{1}{gu} > b > \dfrac{v}{u}$$
If $g > 2 u$, i.e. $g > \sqrt{2n}$, then $v/u$ must be a convergent of the
continued fraction for $b$. Moreover, if we have a bound $a_k < M$ for the elements $a_k$ of this continued fraction, then $gu < (M+ 2) u^2$ so
that $g < \sqrt{(M+2) n}$.

EDIT: In the case $b = \sqrt{3}$, the convergents of the continued fraction that are less than $b$
are $v_k/u_k$ where $u_k$ and $v_k$ both satisfy the recurrence
$a_{k+2} = 4 a_{k+1} - a_k$, with $v_1 = 1$, $v_2 = 5$ while $u_1 = 1$, $u_2 = 3$, and $3 u_k^2 - v_k^2 = 2$. We can then take
$$ \eqalign{g &= \left \lfloor \dfrac{1}{u_k \sqrt{3} - v_k}\right \rfloor = \left \lfloor \dfrac{u_k \sqrt{3} + v_k}{2}\right \rfloor \approx \dfrac{\sqrt{3}-1}{2} (2+\sqrt{3})^k\cr
n &= g u_k \approx \left(\dfrac{\sqrt{3}}{3}-\dfrac{1}{2}\right) (2 + \sqrt{3})^{2k}\cr
\dfrac{g}{\sqrt{n}} &\approx 3^{1/4}}$$

EDIT: A similar calculation for the case $b = \sqrt{7}$ seems to yield
$g/\sqrt{n} \approx (28/9)^{1/4}$