Yes, to each of your questions. A poset is a set $P$ and a partial-order relation defined on the set; when trying to determine if a relation and the set on which it is defined is a poset, you need to check each of the defining properties of a partial order relation with respect to the set. That is true when testing any relationship, not always denoted by $\leq$.
(You need to also show that the relation is antisymmetric, if it is to be a poset.)
In the following, we use the notation "$\,\leq\,$" to represent any partial order relation, not just the narrower/usual meaning of "$\,\leq\,$" ('is less than or equal to'), which also defines a poset on suitable sets. There are many relations which, together with the sets on which they are respectively defined, form a poset. You'll likely encounter some such examples in you studies and/or in problem sets.
A poset consists a set $P$ and a relation on the set. A relation on the set form a poset if and only if:
- $\forall a \in P, \;a\leq a\;$ (reflexity);
- $\forall a, b, c \in P,\;$ whenever $a \leq b$ and $b \leq c$, then this implies $a \leq c\;$ (transitivity);
- $\forall a, b \in P,\;$ whenever $a \leq b$ and $b\leq a$, then this implies $a = b\;$ (antisymmetry).
So given a relation and a set on which it is defined, in order to determine whether you have a poset, you need to confirm that each of these three properties hold for all $a, b, c$ in the set.
Put differently, if there are no $a \in P$ for which reflexivity fails, then the relation reflexive; if there are no $a, b, c \in P$ such that transitivity fails, then the relation is transitive. And if there are no $a, b\in P$ in the set such that antisymmetry fails, then the relation is antisymmetric.