(Although I have pretty much "retired" from Mathoverflow, I will answer this, since the answer is nice but probably not all that well known.)
Theorem. If $X$ is complex normal projective variety, then it is still true that $b_1(X)=2h^1(\mathcal{O}_X)$.
Proof. Let $\pi:\tilde X\to X$ be a desingularization. Since $X$ is normal, the fibres of $\pi$ are connected. Therefore $\pi_*\mathbb{Z}$ (with analytic topology) is connected. It follows that $H^1(X,\mathbb{Z})\to H^1(\tilde X, \mathbb{Z})$ is injective since it can be identified with the edge map for Leray. Therefore the, a priori mixed, Hodge structure on $H^1(X)$ is pure of type $\lbrace (1,0),(0,1)\rbrace$. Consequently $$b_1(X)=2\dim [H^1(X)]^{(0,1)}=2 \dim im[H^1(X,\mathcal{O}_X)\to H^1(\tilde X, \mathcal{O}_{\tilde X}] = 2h^1(\mathcal{O}_X)$$
Regarding your question 2, you can construct an Albanese map $X\to Alb(X)$ to the torus associated to the dual Hodge on $H_1(X)$. The space of $1$-forms on $Alb(X)$ will pullback to a space of the required dimension on $X$.