The exponential sequence does not quite work that way for the algebraic cohomology $H^q(X,\mathcal{O}_{X}^*)$ for $q>1$. Of course the analytic cohomology, $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$, does satisfy the exponential sequence. However, GAGA does not directly apply for $q>1$. Of course it does apply for $q=1$, because both the analytic and algebraic cohomology groups classify invertible sheaves, and GAGA gives an equivalence between these groups. But for $q=2$, one only has that the torsion subgroup of $H^q(X,\mathcal{O}_X^*)$ (and this group is a torsion group in "typical" cases) equals the torsion subgroup of $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$. For a cubic threefold, there is no torsion in $H^3(X^{\text{an}};\mathbb{Z})$. Thus $H^2(X,\mathcal{O}_X^*)$ is zero.

The exponential sequence does not quite work that way for the algebraic cohomology $H^q(X,\mathcal{O}_{X}^*)$ for $q>1$. Of course the analytic cohomology, $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$, does satisfy the exponential sequence. However, GAGA does not directly apply for $q>1$. Of course it does apply for $q=1$, because both the analytic and algebraic cohomology groups classify invertible sheaves, and GAGA gives an equivalence between these groups. But for $q=2$, one only has that the torsion subgroup of $H^q(X,\mathcal{O}_X^*)$ (and this group is a torsion group in "typical" cases) equals the torsion subgroup of $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$. For a cubic threefold, there is no torsion in $H^3(X^{\text{an}};\mathbb{Z})$. Thus $H^2(X,\mathcal{O}_X^*)$ is zero.

Edit. I should make clear that "algebraic cohomology" above really means etale cohomology.