Cohomology computation Let $\Sigma_2$ be the second Hirzebruch surface and $f : \Sigma_2 \to X$ that contracts the exceptional section $E$. Since $E^2=-2$, $X$ has a quadratic singularity at $p := f(E)$. 
If $U'$ is a small ball around $p$, $U = U' \cap X$ and $V = X \backslash \{p\}$, then $X = U \cup V$, $V$ is a line bundle over $\Bbb P^1$ and $U \cap V$ is a $\Bbb C^*$-bundle over $\mathbb P^1$. They are both classified by a number $a \in \mathbb Z$.

Question 1 :  What is $a$ ? (My guess is $a = \pm 2$).

Since $U$ is contractible we get a long exact sequence $$ 0 \to H^1(X) \to H^1(V) = 0 \to H^1(U \cap V) \to  H^2(X) \to H^2(V) \to H^2(U \cap V) \to H^3(X) \to H^3(V) = 0 \to H^3(U \cap V) \to H^4(X) \to 0$$
Hence we get $H^4(X) = \Bbb C, H^1(X) = 0$, and an exact sequence $$ 0 \to H^2(X) \to \Bbb C \to H^2(U \cap V) \to H^3(X) \to 0 $$ 

Question 2 : How to compute $H^2(X)$ and $H^3(X)$ ?

I suspect $H^2(X) = \Bbb C$ and $H^3(X) = 0$.
Moreover I would like to understand :

Question 3 : How to compute the mixed Hodge structure on $X$ associated to the resolution ?

It's clear for $H^0$ and $H^4$. I don't really know what happens for $H^2(X)$. In fact, I read that in such setting, the Hodge structure is never pure, which is really what motivated my question.
 A: Topology of Hirzebruch Surface $\Sigma_2$
Let $Z$ be the zero section and $E$ the $\infty$-section. Let $U=\Sigma_2-E$, $V=\Sigma_2-Z$ be the "neighborhoods" of zero and infinity section respectively, then the projection $U\to Z$ is the total space of line bundle $\mathcal{O}_{\mathbb P^1}(2)$, and $V\to E$ is the dual bundle $\mathcal{O}_{\mathbb P^1}(-2)$. The two line bundles are glued together via $$z\mapsto z^{-1}$$
in fiber direction to get the compact manifold $\Sigma_2$ (See my post here).
The intersection $U\cap V$ is diffeomorphic to $(0,\infty)\times (S^3/\mathbb Z_2)$, where the $\mathbb Z_2$ action on sphere $S^3$ is the antipodal map. Therefore $U\cap V$ has homotopy type $\mathbb R\mathbb P^3$.
By standard topology, $H_1(\mathbb R\mathbb P^3,\mathbb Z)=\mathbb Z/2\mathbb Z$, $H_2(\mathbb R\mathbb P^3,\mathbb Z)=0$. So In the Mayer-Vietoris sequence,
$$0\to H_3(X)\to 0\to \mathbb Z[Z]\oplus \mathbb Z[E]\xrightarrow{\sigma} H_2(X)\xrightarrow{\delta}\mathbb Z/2\mathbb Z\to 0$$
By standard computation on intersection pairing, I claim that $H_2(X)\cong \mathbb Z[F]\oplus\mathbb Z[E]$, with $F$ being a fiber of $\Sigma_2\to Z$, and $\sigma([Z])=[E]+2[F]$. Also, tracing the connecting homomorphism in the level of chain complex, $[F]$ is the class whose image under $\delta$ is the generator of $\mathbb Z/2\mathbb Z$. This answers the first two questions.
Mixed Hodge Structure
Actually, in this case $H^2(X)$ is still pure. To see this, the blowup $f:\Sigma_2\to X$ gives the exact sequence of mixed hodge structures
$$H^1(E)=0\to H^2(X)\xrightarrow{f^{*}}H^2(\Sigma_2)\to H^2(E)\to 0$$
by consider the relative cohomology of the pair $(X,E)$ and uses the fact that $\Sigma_2/E=X$ as CW complex. In particular, $H^2(X)$ includes in a pure Hodge structure, so it is automatically pure.
