What follows is a standard solution using the argument principle. There are many clever approaches that will work, by choosing the right places to use Rouché's theorem and using some inequalities, but this general approach almost never fails.
Argument principle: The argument principle states that over a closed curve, the number of zeros minus the number of poles is the change in argument over the curve divided by $2\pi$. This requires of course that there are no zeros or poles on the boundary. We can write it as $$2\pi(n-m)=\Delta_C \arg(f).$$
To use this, first note that there are no real roots with positive real part since all the coefficients are positive. Since all the coefficients are real, every root in the first quadrant will correspond to a root in the fourth quadrant, so we need only consider the roots in the first quadrant.
Now, notice that there are no roots on the imaginary axis since
$$(ix)^4+8(ix)^3+3(ix)^2+8(ix)+3=x^4-8ix^3-3x^2 +8ix+3=0$$ implies that both $$x^4-3x^2+3=0,\text{ and } 8x-8x^3=0$$ which is impossible.
Consider the contour which is a pizza slice of the right quarter circle or Radius $R$ in the first quadrant. (That is, start at zero, go along the positive real axis until $R$, and then follow the circle or radius $R$ until the imaginary axis, and then return back to the origin) We need only find the change in argument of the polynomial along this curve to know how many zeros are inside. (Since it has no poles.)
Computing the change in argument: We break this down into the three segments.
Real line: On the real line, our polynomial it is always positive, so the change in argument is zero.
Circular arc: On the circular arc, when we take $R$ larger and larger, the first term $z^4$ will dominate so that the change in argument there is the same as it is for $z^4$ in the limit. This gives a change in argument of $2\pi$.
Imaginary axis The imaginary axis is hardest to deal with. However, since the real part is $$x^4-3x^2+3=\left(z^2-\frac{3}{2}\right)^2+\frac{3}{4}$$ which is always positive, we see that it cannot leave the right half plane, and cannot wind around the origin. Hence only the final point and initial point matter, so we see the change in argument is $0$.
Hence as $R\rightarrow \infty$, we see that the change in argument is $2\pi$ so that there is exactly 1 zero. This means that there are two zeros in the right half plane.
Hope that helps,
As I commented, your function $F$ is not well-defined. However, your question still makes sense without it. You can ask, given a multiset of nonzero complex numbers $c_1,\dots,c_n$ invariant under conjugation and $\epsilon>0$, does there exist $\epsilon'>0$ such that for any $r_1,\dots,r_n\in\mathbb{R}$ with $|r_k-|c_k||<\epsilon'$ for each $k$, there exist $d_1,\dots,d_n\in\mathbb{C}$ also invariant under conjugation such that $|d_k|=r_k$ for each $i$ and coefficients of the monic polynomial of with the $d_k$ as roots are within $\epsilon$ of the coefficients of the monic polynomial with the $c_k$ as roots. (The invariance under conjugation condition here corresponds to the requirement that the polynomial has real coefficients.)
The answer to this question is no. For instance, let $n=2$, and $(c_1,c_2)=(i,-i)$ and consider $r_1=1+\epsilon'/2$ and $r_2=1$. Notice then that if $|d_k|=r_k$ for $k=1,2$, then $d_1$ and $d_2$ cannot be conjugates of each other, and so they must both be real in order to be invariant under conjugation. It follows that there are no such $d_1$ and $d_2$ that give rise to a polynomial close to the polynomial $t^2+1$ with roots $c_1$ and $c_2$.
Or, in more concrete terms, there is no polynomial with real coefficients close to $t^2+1$ whose roots have distinct absolute values, since any nearby polynomial has no real roots and hence the roots must be a complex conjugate pair.
Best Answer
It always has infinitely many roots if $A$ or $B$ are not the zero polynomial and $k>0$ since then the function $f(s)$ = $A(s)+e^{sk}B(s)$ is integral of order 1 and finite type $k$ and such have infinitely many zeros unless they are of the type $e^{P(s)}C(s)$, where $P$ has degree 1 and $C$ is a non-zero polynomial; if so, it follows $P(s)$=$ks+r$ and $A(s)$ identically zero. As for continuos dependency, it should follow on any compact set by Hurwitz's theorem.
For $k<0$ same applies just that the type is now $|k|$.