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For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by $p:L\oplus L\rightarrow L$ must be nonzero on one of the summands, say the first one; hence it is the multiplication by a nonzero scalar $\alpha $. Then the map $L\xrightarrow{\ (\alpha ^{-1},0)\ } L\oplus L$ is a section of $p$.

For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by the surjection $p:L\oplus L\rightarrow L$ must be nonzero on one of the summands, say the first one; hence it is the multiplication by a nonzero scalar $\alpha $. Then the map $L\xrightarrow{\ (\alpha ^{-1},0)\ } L\oplus L$ is a section of $p$.

For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by $p:L\oplus L\rightarrow L$ on the first summand is the multiplication by a nonzero scalar $\alpha $; then the map $L\xrightarrow{\ (\alpha ^{-1},0)\ } L\oplus L$ is a section of $p$.

For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by $p:L\oplus L\rightarrow L$ must be nonzero on one of the summands, say the first one; hence it is the multiplication by a nonzero scalar $\alpha $. Then the map $L\xrightarrow{\ (\alpha ^{-1},0)\ } L\oplus L$ is a section of $p$.

No. Consider the associated cohomology exact sequence $$0\rightarrow H^0(L)\rightarrow H^0(L)\oplus H^0(L)\rightarrow H^0(L)\xrightarrow{\ \partial \ }H^1(L)$$ Then $\partial (1)\in H^1(L)=\operatorname{Ext}^1(L,L) $ is the class of the extension given by your exact sequence. But since $\dim H^0(L)=1$, we have $\partial =0$, hence the extension is trivial.

For any line bundle $L$, such exact sequence splits. Indeed the map $L\rightarrow L$ induced by $p:L\oplus L\rightarrow L$ on the first summand is the multiplication by a nonzero scalar $\alpha $; then the map $L\xrightarrow{\ (\alpha ^{-1},0)\ } L\oplus L$ is a section of $p$.