Where does one begin? I can see that the zeros are -5, -3, 0, and 4? Is that correct so far?
Let's use the fact that the graph has zeros at $5,3$ and $-4$. Given the window, we may as well assume these are the only zeros of the polynomial. Note also that we don't have any "flattening" near the zeros, so the zeros must be of multiplicity $1$. Hence, the polynomial is $$p(x)=a(x-5)(x-3)(x+4)$$ for some constant $a$. We can use the fact that $p(0)=3$ to solve for $a$, and then expand if we desire.
No, that is not correct.
The zeros are points where $y = 0$, and it appears to me that none of the points you mentioned are actually zeroes of the function.
This graph has $3$ points where $y = 0$, so by the fundamental theorem of algebra, you know that the polynomial will have degree $3$ (or higher, since there may also be complex roots).
You can use the existence of local minimum and maximum points on the graph to construct a polynomial that will have the same extrema. It is a simple application of differential calculus.