There are a lot of questions here. Let me address Question 3 first:
Let $f$ be a function on some set $A$. When we say that $f$ is defined on $A$, we mean that $f(x)$ exists for every $x$ in $A$. However, just because $f$ is defined on $A$ doesn't mean that it is continuous on $A$: it may be continuous, or it may not be. The textbook intuition of continuity is that $f$ is continuous on $A$ if we can draw its graph without picking up our pencil.
Note, though, that to even talk about continuity on a set, our function has to be defined there first.
Now vector fields are really just a special type of function (one that inputs points and outputs vectors), so all of this applies to vector fields, too.
Before I address Questions 1 and 2, let's first distinguish between a few concepts.
Let $F$ be a vector field defined, continuous, and having continuous partial derivatives on a region $U$ in $\mathbb{R}^3$. Consider the following four properties that $F$ may have:
(1) (Path-independence.) The line integral of $F$ between two points does not depend on the path chosen. That is, if $C_1$ and $C_2$ are paths between $a$ and $b$, then $\int_{C_1} F\cdot dr = \int_{C_2} F\cdot dr.$
(2) (Integrals of closed loops are zero.) The line integral of $F$ around any closed curve is zero. That is, if $C$ is a closed curve, then $\int_C F\cdot dr = 0$.
(3) (Exactness.) There exists a scalar function $f$ with continuous partial derivatives such that $F = \nabla f$.
(4) (Closedness.) $\text{curl }F = 0$.
There are three important facts about these four properties:
Fact 1: Properties (1), (2), and (3) are equivalent.
That is, if $F$ satisfies any one of (1), (2), or (3), then it satisfies the other two. Any of these three properties may be taken as the definition of conservativity.
Fact 2: If $F$ is conservative (i.e. satisfies (1), (2), or (3)), then it also satisfies property (4).
Proof: If $F$ satisfies (3), say, then $\text{curl } F = \text{curl }\nabla f = 0$.
Fact 3: If in addition $F$ is defined on all of $\mathbb{R}^3$ and has continuous partial derivatives, then property (4) implies the first three.
In other words, if $F$ is defined on all of $\mathbb{R}^3$ (and has continuous partial derivatives), then all four concepts coincide.
However: As yoyo mentioned in the comments, property (4) is not in general equivalent to conservativity. The classic example is the vector field on $\mathbb{R}^2 \setminus (0,0)$ given by $F(x,y) = \left(\frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2} \right).$ This vector field has the curious property that $\text{curl }F = 0$, yet does not satisfy any of the three (equivalent) conservativity properties. The reason for this is that $F$ is not defined on all of $\mathbb{R}^2$ because it is undefined at the origin $(0,0)$.
There are similar examples for $\mathbb{R}^3 \setminus (0,0)$ but I can't seem to come up with any right now.
So now we can address Question 2. As I mentioned above, conservativity is not the same as saying that $\text{curl F} = 0$. However, if $F$ is defined on all of $\mathbb{R}^3$, then yes, they coincide.
So why is $\text{curl }F = 0$ such a good property to have? Well, the easy answer is that: (a) it's an easy property to check, and (b) in the event that $F$ is defined on all of $\mathbb{R}^3$, then we get the three (equivalent) conservativity conditions, which I hope you can see are good things to have.
The physical intuition behind conservativity is that it models conservative forces like gravity. In this scheme, $F$ would represent the force and $\int_C F\cdot dr$ would represent the work (energy) of applying $F$ to some object over a path $C$. It is intuitively clear that the work done by gravity (say) over any closed loop is zero: the potential energy hasn't changed any.
Finally, let me address Question 1. Technically speaking, to talk about conservativity, you're right, we don't actually need the partial derivatives of $F$ to be continuous, or even to exist. That is, we can talk about properties (1), (2), and (3) above without the assumption that the partial derivatives of $F$ exist.
However, to talk about property (4), where we take the curl, we need to say that the partial derivatives of $F$ exist. Do we need to assume in addition that the partial derivatives are continuous? My guess is probably not (someone please correct me if I'm wrong). But assuming that the partial derivatives are continuous is certainly a nice simplifying assumption to have.