Let $f=f(t),g=g(t) \in k[t]$, whrer $k$ is a field of characteristic zero.
Assume that:
(i) $k(f,g)=k(t)$. Therefore, there exist $P(X,Y),Q(X,Y) \in k[X,Y]$, such that $t=\frac{P(f,g)}{Q(f,g))}$.
(ii) $f=\sum_{i=1}^{n}a_it^i$, $g=\sum_{j=1}^{m}b_jt^j$, where $a_i,b_j \in k$ and $a_1b_1 \neq 0$. Therefore, $f'(0)=a_1$, $g'(0)=b_1$, $f(0)=0$, $g(0)=0$.
By questions 1 and 2, there exist $a,b,c \in k$ such that $\gcd(f-a,g-b)=t-c$.
When is it possible to find $a,b \in k$ such that $\gcd(f-a,g-b)=t$? namely, when can we can take $c=0$?
According to the answer in 2, $c=0$ is possible, if there exist no $d \in k$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$. Here, $f(c)=0$, so if $d$ is another root of $f$, we must have $Q(0,g(d))=Q(f(d),g(d)) \neq 0$. However, I do not know what is the form of such $Q$; if we knew that $Q(0,g(d) \neq 0$, then we were done).
Remark: Here $t$ divides both $f$ and $g$, but perhaps there exist additional common divisors of $f$ and $g$. Notice that $t^2$ does not divide $f$ nor $g$, since otherwise, $f'(0)=g'(0)=0$, but here $f'(0)=a_1 \neq 0$ and $g'(0)=b_1 \neq 0$.
Thank you very much!