The answer should be $\ce{-CH2OH}$ in the first part. And, for the double bond, we have according to this Wikipedia about the Cahn Ingold Prelog rules,
If an atom A is double-bonded to an atom B, A is treated as being
singly bonded to two atoms: B and a "ghost atom" that is a duplicate
of B (has the same atomic number) but is not attached to anything
except A.
But still, we need to arrange according to the decreasing atomic number as according to the same article,
a list is made for each group of the atoms bonded to the one directly
attached to the stereocenter. Each list is arranged in order of
decreasing atomic number. Then the lists are compared atom by atom; at
the earliest difference, the group containing the atom of higher
atomic number receives higher priority.
We will have the list as $\ce{O, H, H}$, for $\ce{−CH2OH}$ and as $\ce{C, C, H}$ in $\ce{-CH=CH2}$. On comparing the first difference, it is easily seen that $\ce{O}$ should be given the priority.
In the second part too, we need to consider the first difference which arises, and in this way, you can work out the solution.