Let $k$ be an algebraically closed field, and $K\supset k$ be an algebraically closed extension. Let $a\in K^n$ be a tuple, we call $a^\prime\in k^n$ a specialization of $a$ if for any $f(X)\in k[X]$ with $f(a)=0$ we have $f(a^\prime) = 0$.
Now, let $f_i(X,Y)\in k[X,Y]$ for $i=1,...,n$. Assume also that there is a tuple $(a,b)$ such that $f_i(a,b) = 0$ and $a$ specializes to $a^\prime$.
1) Is it true that the system of polynomial equations $f_i(a^\prime,Y) = 0$ has a solution in $k^m$?
2) Is it true that $f_i(a^\prime,b) = 0$ for all $i$?
Part 2 the answer is no, and it can be seen in the simplest example where $k = \mathbb{C}$ and $K = \overline{\mathbb{C}(t)}$, and $n=1$.
Let $(a,b) = (t,\sqrt{t}) \in K$. Then $F(X,Y)= X - Y^2$ vanishes at $(a,b)$ but if we specialize $t$ to any complex number $a'$ then $F(a',b) = a'-t \ne 0$.
Edit: Part 1 the answer is also no, and can again be seen in the same example of fields as above with $n=1$.
This time take $(a,b) = (t,1/t)$ and let $F(X,Y) = XY-1$. Clearly $F(a,b) = 0$, but if we specialize $t$ to $0$ then $F(0,Y) = 0$ has no solutions.