Consider an elliptic curve $X$. The $\mathcal{O}_X$-module $\mathcal{O}_X$ itself is free, so in particular projective. Thus, $Ext^1(\mathcal{O}_X, L)=0$ for all line bundles $L$ over $X$ by Exercise 2.5.2 in Weibel: An Introduction to Homological Algebra.
On the other hand, $Ext^1(\mathcal{O}_X, L)=H^1(\mathcal{O}_X^*, L)$ by the exercise on the bottom of page 1 in these handwritten notes titled Extensions of line bundles. Now, using $\mathcal{O}_X^*=\mathcal{O}_X$ and applying Serre duality gives: $$ Ext^1(\mathcal{O}_X, L) = H^1(\mathcal{O}_X^* \otimes L) = H^0(K_X \otimes L^*)^* = H^0(L^*)^*. $$ The term $H^0(L^*)^*$ can be non-zero, namely if $L^*$ has non-negative degree. This contradicts the first paragraph.
Question: Which is true? Is $Ext^1(\mathcal{O}_X, L)=0$ for all line bundles $L$ over $X$?
Context for the question: I'm trying to get an explicit description of an extension of a degree $1$ line bundle by the trivial line bundle over an elliptic curve. In order to get this explicit description, I'm trying to trace the proof of Theorem 3.4.3 in Weibel: An Introduction to Homological Algebra, where it is used that certain $Ext$ groups vanish.
On a ringed space $(X, \mathcal{O}_X)$ with an $\mathcal{O}_X$-module $\mathcal{G}$ one can show in general (by writing out the definitions) that $$\operatorname{Ext}_X^i(\mathcal{O}_X, \mathcal{G}) \cong H^i(X, \mathcal{G}).$$ See for example Vakil, #30.2.C. In particular, this does not need to vanish and you will know many examples where it doesn't vanish.
This also shows that $\mathcal{O}_X$ in general is not projective (despite it being free in the sense of Hartshorne) since otherwise we would have $\operatorname{Ext}^1_X(\mathcal{O}_X, -) = 0$ by the usual homological algebra argument. (A more elementary formulation is as written in Qiaochu Yuan's comment: The global sections functor would be exact.)
So the error here is that (Hartshorne's) freeness does not imply projectivity, as one might be used to from ordinary modules over rings.