**Question 1:** Sure, this is true. Another reference (beyond what Donu pointed out) is chapter 8 of Claire Voisin's *Hodge theory of ...*. But the point is $\Omega_X^{\bullet}(\text{log} D)$ is quasi-isomorphic to the pushforward of a resolution of $\mathbb{C}$ on $X \setminus D$. By the way, this holds not just for smooth $D$, but also for normal crossing $D$.

**Question 2:** I don't think you have maps of complexes as you describe. For example, why do we have the map $\Omega_V^1(\text{log} D) \to \Omega^{\bullet}_V(\text{log} D)$? If I had a map of complexes, then the image of $d : \Omega_V^0(\text{log} D) \to \Omega_V^1(\text{log} D)$ would be zero (ie, the diagram would commute).

**EDIT:** Whoops, it looks like Donu beat me to this in the comments.

**Revised Question 2:** I don't see why this should hold in general. However, if you write down the relevant spectral sequence, and enough terms vanish (maybe the spectral sequence degenerates), you can be ok.

**EDIT (Response to the comment below):** No, there isn't a map in general. Even for a projective variety and $D = 0$, we only have an $E^1$ degeneration of the spectral sequence. This means that in some sense, $H^2(X, O_X)$, $H^1(X, \Omega_X^1)$ and $H^0(X, \Omega_X^2)$ make up ${H}^2(X, \mathbb{C})$ (there is a filtration of the latter such that these terms make up the filtration). But we have maps:

$$H^2(X, \mathbb{C}) \to H^2(X, O_X), \text{ and } H^0(X, \Omega_X^2) \to H^2(X, \mathbb{C}).$$

There isn't going to be a map to the $H^1(X, \Omega_X^1)$ term in general, unless for some reason $H^2(X, O_X) = 0$ (in the non-compact/non-Kahler setting, things get more complicated as Donu mentioned above). Anyways, if you read a little about spectral sequences and do a couple examples from that perspective, I'm sure you'll see what's going on.