Recall what a subspace must satisfy : it must be closed under addition and scalar multiplication, and contain the zero element (and must be a subset of the vector space you are checking it is a subspace of, but this is clear here).
In none of the conditions does $p(2x)$ come into play. You should explain from where the idea that $p(2x) = 2p(x)$ must be true, came to you. The check performed is incorrect.
Note that $p(2x)$ is not the scalar multiplication of $2$ with $p(x)$ : it is the polynomial $p$, applied to the input $2x$. This is something that could differ wildly from $p$ : for example, if $p(x) = 3x^2 + 5x + 6$ then $p(2x) = 12x^2 + 10x + 6$, which is very different from $p(x)$, and will not be $2p(x)$ in general.
For the first example, remember the three points : closed under addition, closed under scalar multiplication, and zero is a member.
Is it closed under addition? $ax^2+bx^2 = (a+b)x^2$.
Is it closed under scalar multiplication? $c(ax^2) = (ca)x^2$.
Does it contain zero? $0 = 0(ax^2)$.
Hence, it is a subspace.
The second set does not contain zero, since $a+x^2$ has degree two always, while $0$ has no degree. So, $0 = a+x^2$ cannot happen, hence the second is not a vector space.