It's important to be precise in your context, here. If you are discussing a poset $P$, and wondering if $P$ has a least upper bound in the poset $P$, then you are wondering precisely if $P$ has a maximum element.
However, if you were discussing a subset of $P$, say $A$, and wondering if $A$ has a least upper bound in the poset $P$, then that's a different matter. If $A$ has a maximum element, then it will readily be the least upper bound of $A$. However, if $A$ has no maximum element, but only maximal elements--such as $A=\{a,b\}$ with $a,b$ incomparable--then the least upper bound of $A$ (if it exists) will necessarily be a member of $P$ that is not in $A$.
In your particular example with $A=\{4,5,7\}$, note that $4$ is incomparable with both $5$ and $7$, and that $5$ is greater than $7$. Hence, any upper bound of $A$ need only be an upper bound of $4$ and $5$, and any lower bound of $A$ need only be a lower bound of $4$ and $7$. The set of all upper bounds of $A$ is $\{1,2,3\}$ and this set has a least element, namely $3$, so $\sup A=3$. The only lower bound of $A$ is $8$, so $\inf A=8$.
Note that in general, $\sup A$ (if it exists) is the minimum element of the set of upper bounds of $A$, and $\inf A$ (if it exists) is the maximum element of the set of lower bounds of $A$. If $A$ has no upper (lower) bounds, then $\sup A$ ($\inf A$) doesn't exist. For example, if we didn't have the node $8$ in your example $W$, then $\inf A$ wouldn't exist. If the set of upper (lower) bounds of $A$ is non-empty, but has no minimum (maximum) element, then $\sup A$ ($\inf A$) doesn't exist. For example, suppose we were to add another node, $9$, to your example, on the same level as $8$, with $9\prec 6$, $9\prec 7$, and with $8$ and $9$ incomparable. In that case, the set of lower bounds of $A$ would be $\{8,9\}$, but since $8$ and $9$ are incomparable, that set has no maximum element, meaning $\inf A$ wouldn't exist.
In this cases double inclusion is the way: let $x \in f(f^{-1}(X))$, then there is $a \in f^{-1}(X)$ such that $f(a) = x$. By definition of the pre image, $x \in X$. This gives $$f(f^{-1}(X)) \subset X.$$
Note that we didn't use that the function is surjective to prove this inclusion, meaning that it holds in general.
Now let $x \in X$. Being $f$ surjective, we can find $a \in A$ such that $f(a) = x$. This gives $a \in f^{-1}(X)$ and hence $x = f(a) \in f(f^{-1}(X))$. This shows $$X \subset f(f^{-1}(X)).$$
Best Answer
I agree with your statements 2 and 3.
For statement 1, there might be a ``trap".
Well, here are two things we know about the maximum cardinality of $f[A]$, say $|f[A]|_{MAX}$.
Since we don't know which is larger between $|A|$ and $|B|$, it is better to say $|f[A]|_{MAX} = \min\left\{ |A|, |B|\right\}$