Step 1: Instead of citing the kernel-of-a-linear-map fact, test each set directly against the three subspace axioms: contains 0, closed under addition, closed under scalar multiplication.
Step 2: $W_1$: the zero polynomial gives $0(1/2) = 0$, a rational number, so $0 \notin W_1$. Since every subspace must contain $0$, $W_1$ fails immediately.
Step 3: $W_2$: take $p(x) = 1$ and $q(x) = 2x$; both satisfy $p(1/2)=1$ and $q(1/2)=1$, so $p,q \in W_2$. But $(p+q)(1/2) = 1+1 = 2 \neq 1$, so $p+q \notin W_2$, meaning closure under addition fails. $W_2$ is not a subspace.
Step 4: $W_3$: let $p,q \in W_3$, so $p(1/2)=p(1)$ and $q(1/2)=q(1)$, and let $c$ be any scalar. Writing $s=p+q$ and $t=cp$: \[s(1/2) = p(1/2)+q(1/2) = p(1)+q(1) = s(1), \qquad t(1/2) = c\,p(1/2) = c\,p(1) = t(1)\] so $s,t \in W_3$. The zero polynomial trivially satisfies $0(1/2)=0=0(1)$, so $0 \in W_3$. All three axioms hold, so $W_3$ is a subspace.
Step 5: $W_4$: take $p(x) = x$, so $p'(x) = 1$ everywhere, giving $p'(1/2) = 1$, so $p \in W_4$. But $p+p = 2x$ has derivative $2$, so $(p+p)'(1/2) = 2 \neq 1$, meaning $p+p \notin W_4$. Closure under addition fails, so $W_4$ is not a subspace.
Step 6: Direct verification confirms only $W_3$ is a subspace. \[\boxed{W_3 = \{p \in V : p(1/2) = p(1)\}}\]