= Solution
A <holomorphic line bundle> is a complex rank-one bundle locally trivialized over <holomorphic coordinate> opens, with nowhere-zero <holomorphic> transition functions. A local <holomorphic section> has a <holomorphic> coefficient in a <holomorphic local frame>. Let $h$ be a smooth positive <Hermitian metric> on its fibres. A <Chern connection> is a connection compatible with $h$ whose $(0,1)$ part is the bundle's <Dolbeault operator>.
Fix the Hermitian-product convention linear in the first variable and a nonvanishing <holomorphic local frame> $e$. Put $h_e=h(e,e)>0$ and write $\nabla e=Ae$. Compatibility with the <holomorphic> structure forces $A^{0,1}=0$. Metric compatibility gives
$$
dh_e=(A+\bar A)h_e,
$$
so taking its $(1,0)$ part determines $A$ uniquely:
$$
\boxed{A=\partial\log h_e.}
$$
Define the connection by this formula. To check it exists globally, change frame to $e'=ge$ for a nonzero <holomorphic function> $g$. Then $h_{e'}=|g|^2h_e$ and
$$
\partial\log h_{e'}=\partial\log h_e+g^{-1}dg,
$$
which is exactly the connection transformation law obtained from $\nabla(ge)=dg\,e+g\nabla e$. Hence the local connections glue. The formula and its conjugate give metric compatibility, and $A^{0,1}=0$ gives the required <holomorphic> structure. Since either condition already forced this unique formula in every frame, existence and uniqueness are proved.
For a <line bundle>, the <curvature form of a connection> is $F=\nabla^2=dA$, as the scalar one-form satisfies $A\wedge A=0$. The <local formula for the Chern connection on a line bundle> therefore gives
$$
\boxed{F_h=\bar\partial\partial\log h_e=-\partial\bar\partial\log h_e.}
$$
This formula uses a nowhere-zero local <holomorphic section>; arbitrary sections with zeros are not frames there. It has type $(1,1)$, is $d$-closed and satisfies $\bar F_h=-F_h$, so $iF_h$ is a real $(1,1)$-form. The transformation calculation makes its frame independence explicit.
For two metrics, their ratio $r=h_1(e,e)/h_2(e,e)$ is a global positive smooth function, since the common frame factors cancel. The <curvature difference of two Chern connections> is
$$
F_{h_1}-F_{h_2}=\bar\partial\partial\log r,
\qquad iF_{h_1}-iF_{h_2}=\bar\partial(i\partial\log r).
$$
Each curvature is $\bar\partial$-closed, and the difference has a global $(1,0)$ primitive for $\bar\partial$. Thus
$$
\boxed{[iF_{h_1}]=[iF_{h_2}]\ \text{in }H^{1,1}(X).}
$$
This is equality in <Dolbeault cohomology>, rather than merely equality in <de Rham cohomology>; the displayed primitive proves precisely that distinction.
Finally let $s$ be a nowhere-zero smooth section of $\hat L$. In this global smooth trivialization write $\bar\partial_{\hat L}s=a\otimes s$ with a global $(0,1)$-form $a$. The <holomorphic> structure has $\bar\partial_{\hat L}^2=0$. Its Leibniz rule gives
$$
0=\bar\partial_{\hat L}(a\otimes s)=(\bar\partial a-a\wedge a)\otimes s=(\bar\partial a)\otimes s,
$$
since a scalar one-form wedges with itself to zero. The hypothesis $H^{0,1}(X)=0$ makes $a=\bar\partial f$ for a global smooth complex function $f$. Then
$$
\bar\partial_{\hat L}(e^{-f}s)=e^{-f}(a-\bar\partial f)\otimes s=0.
$$
The exponential never vanishes. Therefore \b[$e^{-f}s$ is a nowhere-zero <holomorphic section>], proving <smoothly trivial holomorphic line bundles when H01 vanishes> constructively, without a <compactness> assumption.
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