Tangent bundle of a compact two-dimensional manifold 
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*For which (connected) two dimensional compact manifold M, oriented or not, the tangent bundle TM is trivial?

*For which of these manifolds the complexified tangent bundle $T^\mathbb{C}M = TM\otimes \mathbb{C}$ is trivial?

 A: I have an almost complete answer. I start with a summary of the different cases. In all the cases $M$ is assumed to be a closed 2-manifold.

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*$M$ is orientable. Then $TM$ is trivial if and only if $M$ is the torus (genus 1).
Proof: Euler characteristic $+$ Poincaré-Hopf theorem.


*$M$ is orientable. Then $TM \otimes \mathbb{C}$ is trivial if and only if $M$ is the torus (genus 1).
Proof: $c_1$ is equal to the Euler characteristic times 2.


*$M$ is non-orientable. Then $TM$ is not trivial. The trivial bundles are orientable.


*$M$ is non-orientable. If $M$ is not the Klein bottle, then $TM \otimes \mathbb{C}$ is not trivial.
Proof: Consider a covering of $M$ with an oriented manifold, and pull the bundle back.
Remaining case: $TM \otimes \mathbb{C}$ of the Klein bottle. I don't know whether it is trivial or not.
Details:
Let's start with the closed orientable manifolds. In this case
$TM$ is trivial if and only if $TM \otimes \mathbb{C}$ is trivial if and only if $M$ is the torus $T^2$.
Proof: Let $M$ be the $k$-fold connected sum of the torus $T^2$. The Euler characteristic of $M$ is $\chi(M)=2-2k$. By Poincaré-Hopf theorem the Euler characteristic of a closed orientable manifold is equal to the Euler number of its tangent bundle: $\chi(M)=e(TM)$. Here $e(TM)$ is defined as the sum of the indices of any generic tangent vector field, i.e. any continuous section of the tangent bundle. Therefore $TM$ admits a nowhere zero section if and only if $k=1$. Hence $TM$ is nontrivial if $k \neq 0$, and we know that the tangent bundle of the torus is trivial ($T^2=S^1 \times S^1$ and $TS^1$ is trivial.)
The complexified bundle is $TM \otimes \mathbb{C} \cong TM \oplus TM$. $TM$ can be endowed with a complex structure, whose first Chern class number $c_1 (TM)[M]=e(TM)$. Therefore $c_1(TM \oplus TM)=2 \cdot c_1(TM) $ is zero if and only if $k=1$, i.e. when $M$ is the torus.
The tangent bundle $TM$ of a closed nonorientable manifold $M$ is nonorientable. Therefore it cannot be trivial. Its first Stiefel-Whitney class $w_1(TM)$ is nonzero.
In the case of the complexified tangent bunde $TM \otimes \mathbb{C}$ of a nonorientable manifold $M$ the first Stiefel-Whitney class of the complexification does not help. Indeed, it is $w_1(TM \otimes \mathbb{C})=w_1(TM \oplus TM)=2 \cdot w_1(TM)=0$, since $w_1(TM) \in H^1(M, \mathbb{Z}_2)$. (According to the fact that every complex bundle is oriented.)
Instead, consider a $k$-fold covering $p: N \to M$ of the non-orientable manifold $M$ with an oriented one $N$. Such a positive integer $k$ and covering $p$ always exists. Then $p^*(TM) \cong TN$ and $p^*(TM \otimes \mathbb{C}) \cong TN \otimes \mathbb{C} $. Therefore if $TM \otimes \mathbb{C}$ is trivial, then $TN \otimes \mathbb{C}$ is trivial as well, which implies that $N$ is the torus.
On the other hand $k \cdot \chi(M)= \chi(N)$ is zero if $N$ is the torus. Hence the triviality of the bundle $TM \otimes \mathbb{C}$ implies that $\chi(M)=0$, that is, $M$ is the Klein bottle.
We proved that $TM \otimes \mathbb{C}$ is nontrivial, if the nonorientable manifold $M$ is not the Klein bottle. I don't know that the complexification of the tangent bundle $TK \otimes \mathbb{C}$ of the Klein bottle $K$ is trivial or not.
