Assume $F$ is a field. Then for each $f,g\in F[x]$ with greatest common divisor $(f,g)=d$ by Bezout's identity $uf+vg=d$ for some $u,v\in F[x]$.
How can we see that $(u,v)=1$?
Furthermore, if for some $h,r\in F[x]$ we have $hf+rg=1$, how can we see that $(f,g)=1$?
And lastly, if $g$ is not associated to $f$, how can we find unique Bezout coefficients $u,v$ such that $\deg u\leq\deg g-\deg (f,g)-1$ and $\deg v\leq\deg f-\deg (f,g)-1$?
Are those statements true in any Bezout ring?
The last part first, because the first part follows from it.
If $d|f$ and $d|g$, then $d|hf + rg$, so $d|1$. Hence $(f,g) = 1$
Now for the first part: if $(u,v) = e > 1$, then $e | uf + vg = d $, so $\frac{u}{e}f + \frac{v}{e}g = \frac{d}{e}$. So by the above, we must have $(f,g) | \frac{d}{e}$
If $e|u$, $e|v$ then $e|d$
$\therefore e|f$, $e|g$
$\therefore e^2|d$
$\therefore e^2|f$, $e^2|g$
$e^n|d \ \forall n$, contradiction
For your second question suppose $e|f$, $e|g$ then $e|1$
Both argument do work in any Bezout ring, there are no requirements really.
$\begin{eqnarray} {\bf Hint}\ \ \ (a,b)\mid a,b &&\Rightarrow\ (a,b)\mid ua+vb\\ (u,v)\mid u,v\! &&\Rightarrow\ (u,v)\mid ua+vb\end{eqnarray}$
Hence applying the above to $\ u\dfrac{f}d + v \dfrac{g}d\, =\, \color{#c00}1\ $ yields $\,(u,v)\mid \color{#c00}1.\,$
You can find the Bezout coefficients using the Extended Euclidean algorithm.
The divisibility inference holds true in any ring, i.e, any common divisor of linear form is necessarily greatest (i.e. divisible by all common divisors), see here for further discussion.
