I came up with a sketch of something that looks like it works for question 2. It uses the main idea I mentioned in my comment to answer your first question: namely real and complex line bundles correspond to elements of cohomology groups via the isomorphisms
$$ w_1 \colon Prin_{O(1)}(X) \cong H^1(X;\mathbb{Z}/2)\ \ \text{ and }\ \ c_1 \colon Prin_{U(1)}(X) \cong H^1(X;\mathbb{Z})$$
resulting from the homotopy equivalences $BO(1)\simeq K(\mathbb{Z}/2, 1)$ and $BU(1) \simeq K(\mathbb{Z},2)$.
(I haven't solved 3 yet)
To answer question $2$, the meat is in showing that for any $X$ the morphism $\beta\colon H^1(X;\mathbb{Z}/2) \to H^2(X;\mathbb{Z})$ is induced by complexification of bundles.
Edit: Faster proof
The Bockstein $\beta\colon H^1(-;\mathbb{Z}/2) \to H^2(-;\mathbb{Z})$ corresponds to a homotopy class of functions $K(\mathbb{Z}/2,1) \to K(\mathbb{Z}, 2)$, and since this set of homotopy classes is $H^2(\mathbb{RP}^\infty;\mathbb{Z}) \cong \mathbb{Z}/2$, this map is either null-homotopic or in the unique class of maps which are not. To show it's not null-homotopic it suffices to come up with a space for which $\beta$ is non-zero. Indeed you can show that $H^1(\mathbb{RP}^\infty;\mathbb{Z}) = 0$ and $H^1(\mathbb{RP}^\infty;\mathbb{Z}/2)\cong \mathbb{Z}/2$ so $\beta \colon H^1(\mathbb{RP}^\infty;\mathbb{Z}/2) \to H^2(\mathbb{RP}^\infty;\mathbb{Z})$ is non-zero by the Bockstein exact sequence for $\mathbb{Z} \stackrel{\cdot 2}{\rightarrow} \mathbb{Z} \to \mathbb{Z}/2$. Considering that $K(\mathbb{Z}/2,1) \simeq BO(1)$ and $K(\mathbb{Z}, 2) \simeq BU(1)$, and that the complexification map $c\colon BO(1) \to BU(1)$ is not null-homotopic, it follows that $c$ represents $\beta$. See the bottom of my post for the conclusion of the argument.
My original proof with more theory and constructions and is more informative:
First we need some
General theory.
Given a short exact sequence of topological groups $H\to G \to G/H$ such that the quotient map is a fibre bundle, it can be extended to a "delooping" sequence
$$ H \to G \to G/H \stackrel{\delta}{\to} BH \to BG $$
where each sequence of three spaces is equivalent to a fibration; moreover if $H$ is normal then the delooping sequence also extends one more step with a map $BG \to B(G/H)$. The map which is not given by functoriality is the connecting map $\delta\colon G/H \to BH$. To describe this map we use the fact that there is a model of $BH$ given by
$$ BH \simeq EG/H \cong EG \times_G (G/H) $$
where $X\times_G Y$ is the "Borel Construction" or "Balanced Product" (see for example these notes, in particular Section 3, Proposition 8.3, and Theorem 11.3.) In words, there is a model of $BH$ constructed by starting with a model of $EG$ and replacing the fibres with copies of $G/H$; the result is a fibre bundle $BH \to BG$ and the connecting map $\delta\colon G/H \to BH$ is represented by the inclusion of one of these fibres.
If $G$ and $H$ are moreover abelian, then $BG$, $BH$ and $B(G/H)$ are all groups as well, and even abelian themselves, so they too have classifying spaces which are also abelian groups. In this case the delooping sequence extends arbitrarily far:
$$ H \to G \to G/H \stackrel{\delta}{\to} BH \to BG \to B(G/H) \stackrel{B\delta}{\to} B^2H \to B^2G \to B^2(G/H) \stackrel{B^2\delta}{\to} \dots $$
Suppose finally that we are in the extra-special case that $G$ and $H$ are even discrete. Since $B^nA \simeq K(A, n)$ for any discrete abelian group $A$, given any $X$ the homotopy classes of maps into the delooping sequence give the usual long exact sequence of cohomology groups
$$ H^0(X;H) \to H^0(X;G) \to H^0(X;G/H) \stackrel{\beta}{\to} H^1(X;H) \to H^1(X;G) \to H^1(X;G/H) \stackrel{\beta}{\to} H^2(X;H) \to \dots$$
Our case.
We are considering $\beta\colon H^1(X;\mathbb{Z}/2) \to H^2(X;\mathbb{Z})$ coming from the short exact sequence $\mathbb{Z} \stackrel{\cdot 2}{\rightarrow} \mathbb{Z} \to \mathbb{Z}/2$. Our standard model of $E\mathbb{Z}$ will be $\mathbb{R}$ with the usual $\mathbb{Z}$ action, so that $B\mathbb{Z} \cong S^1 \cong U(1)$. To model the map $B(\cdot 2)$ we construct a model of $B(2\mathbb{Z})$ as above to get $\mathbb{R} \times_\mathbb{Z}(\mathbb{Z}/2)$, which takes all of the $\mathbb{Z}$ fibres in the quotient map $\mathbb{R} \to S^1$ and replaces them with copies of $\mathbb{Z}/2$. This results in a connected double cover of $S^1$, since for any $[r,\epsilon]\in \mathbb{R} \times_\mathbb{Z}(\mathbb{Z}/2)$ the path sending $t$ to $[r + t,\epsilon]$ is a path from $[r,\epsilon]$ to $[r+1,\epsilon] = [r,\epsilon + 1]$. Therefore it is equivalent to the unique non-trivial double-cover $S^1 \to S^1$, and it follows that the connecting map $\mathbb{Z}/2 \to B\mathbb{Z}$ in the delooping sequence is actually modelled by the standard inclusion $\mathbb{Z}/2 \hookrightarrow S^1$, which is also known as the "complexification" map $\kappa\colon O(1) \to U(1)$. Therefore the operation $\beta\colon H^n(X;\mathbb{Z}/2) \to H^{n+1}(X; \mathbb{Z})$ is induced by $B^n(\kappa)$. In particular when $n=1$ we get the map $BO(1) \to BU(1)$ which sends a real line bundle to its complexification, or more specifically if $V \to X$ is classified by $f\colon X\to BO(1)$ then $B\kappa \circ f$ classifies $V\otimes \mathbb{C}$ (I think this is worked out in Mirura-Toda somewhere). We end up with the following diagram for all $X$
$\require{AMScd}$
\begin{CD}
Prin_{O(1)}(X) @>\otimes\mathbb{C}>> Prin_{U(1)}(X)\\
@V w_1 V \cong V= @V c_1 V \cong V\\
H^1(X;\mathbb{Z}/2) @>>\beta> H^2(X;\mathbb{Z})
\end{CD}
which expresses what you're trying to show for line bundles.
If $E$ is not necessarily a line bundle, but instead has rank $n\geq 1$, replace $E$ with its determinant line bundle $\Lambda^n E$. Then it's known that $w_1(\Lambda^n E) = w_1(E)$ and $c_1(\Lambda^n E\otimes \mathbb{C}) = c_1(\Lambda^n (E\otimes \mathbb{C})) = c_1(E\otimes \mathbb{C})$ (the proof I know uses the splitting principle to come up with a formula in terms of elementary symmetric polynomials, I'm not sure of a source off the top of my head).
