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I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2-plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$.

Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map

$\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$.

Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2-plane $\Pi$.

Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$.

Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable.

As for the second question, orientability of the total space is of course a necessary condition for it to have a complex structure.

Assuming my construction above is meaningful, if one knows $\pi_1(\mathrm{Diff}^+(F))$, he's able to tell apart the topological types of bundles on a given $S$ for which the orientability obstruction is not present.

(I know I'm being quite vague, but) ...then one may try to look for an (existing?) topological classification of holomorphic (or maybe just algebraic) bundles, to see whether each topological class is in fact realized by a a holomorphic/agebraic example...

I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2-plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$.

Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map

$\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$.

Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2-plane $\Pi$.

Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$.

Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable.

As for the second question, orientability of the total space is of course a necessary condition for it to have a complex structure.

Assuming my construction above is meaningful, if one knows $\pi_1(\mathrm{Diff}^+(F))$, he's able to tell apart the topological types of bundles on a given $S$ for which the orientability obstruction is not present.

(I know I'm being quite vague, but) ...then one may try to look for an (existing?) topological classification of holomorphic (or maybe just algebraic) bundles, for various complex structures on $S$ and $F$, to see whether each topological class is in fact realized by a a holomorphic/agebraic example...

I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2-plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$.

Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map

$\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$.

Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2-plane $\Pi$.

Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$.

Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable.

As for the second question, orientability of the total space is of course a necessary condition for it to have a complex structure.

I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2-plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$.

Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map

$\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$.

Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2-plane $\Pi$.

Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$.

Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable.

As for the second question, orientability of the total space is of course a necessary condition for it to have a complex structure.

Assuming my construction above is meaningful, if one knows $\pi_1(\mathrm{Diff}^+(F))$, he's able to tell apart the topological types of bundles on a given $S$ for which the orientability obstruction is not present.

(I know I'm being quite vague, but) ...then one may try to look for an (existing?) topological classification of holomorphic (or maybe just algebraic) bundles, to see whether each topological class is in fact realized by a a holomorphic/agebraic example...

I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2-plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$.

Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map

$\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$.

Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2-plane $\Pi$.

Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$.

Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable.

I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2-plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$.

Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map

$\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$.

Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2-plane $\Pi$.

Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$.

Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable.

As for the second question, orientability of the total space is of course a necessary condition for it to have a complex structure.