Let $p:S^3 \to S^2$ be the Hopf fibration which is a result of the standard action of $S^1$ on $S^3$.



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> Is there a $2$ dimensional vector bundle $\tilde{p}:E \to S^2$ such that $S^3\subset E$ and $\tilde{p}\_{|S^3}=p$?
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> Is there a vector bundle $E$ as above with the following extra condition:The total space $E$ can be acted by $S^1$ with linear isomorphism and this action would be the extension of the standard action of $S^1$ on $S^3$?
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**Note:** One can ask the same question for such extension of an arbitrary principal bundle to an equivariant vector bundle of arbitrary dimension.



Extension of Hopf fiber bundle to (an equivariant) 2 dimensional vector bundle
I think both of your questions can by answered positive. Since the Hopf fibration is an $S^1$ bundle it comes as a sphere bundle of complex line bundle over $S^2$. From the Gysin sequence the first Chern class of that line bundle must be a generator of $H^2(S^2;\mathbb Z)$ which determines the isomorphism type of the line bundle.


Moreover a circle action on $S^2$ can be lifted to a line bundle if and only if the first Chern class possess an equivariant extension in $H\_{S^1}(S^2;\mathbb Z)$ which here is always true.


**Edit:** I would like to remark that lifting actions from the base space to a vector bundle is in general a very difficult problem (which is solved for complex line bundles) but unknown (as far as I know) for complex vector bundle of rank higher than 2. There are some results for oriented vector bundles over spheres.


