Here is a positive answer to Q1.

Step 1: Without loss of generality, each $X_i$ is symmetrically distributed. Indeed, consider an independent copy $X_i'$ of the family $X_i$ with the same joint distribution. Then
$$
E\left|\sum_i X_i\right|^p\le E\left|\sum_i (X_i-X_i')\right|^p
$$
(only the mean $0$ property and the convexity of $x\mapsto |x|^p$ are used here) while
$$
E|X_i-X_i'|^p\le 2^p E|X_i|^p
$$
and $X_i-X_i'$ are still pairwise independent.

Now choose $Q>0$ so that $Q^{2-p}=2$. Clearly, $Q\ge 2$. Define $X_{i,k}=X_i$ if $Q^k\le|X_i|< Q^{k+1}$ and $0$ otherwise ($k\in\mathbb Z$). Also denote by $Y_{\ell,k}$ the characteristic function of the event $Q^{\ell+k}\le\left|\sum_i X_{i,k}\right|<Q^{\ell+k+1}$ ($\ell\in\mathbb Z$).

Step 2: The key estimates:
When $\ell\ge 0$, we have
$$
E\left[Y_{\ell,k}\left|\sum_i X_{i,k}\right|^p\right]\le Q^{(p-2)(\ell+k)}E\left|\sum_i X_{i,k}\right|^2
\\
=Q^{(p-2)(\ell+k)}\sum_i E\left|X_{i,k}\right|^2\le 2Q^{(p-2)\ell}\sum_i E|X_{i,k}|^p=2^{1-\ell}\sum_i E|X_{i,k}|^p
$$
because $Q^{(p-2)k}|x|^2\le 2|x|^p$ when $x=0$ or $Q^k\le|x|<Q^{k+1}$.

If $\ell<0$, then
$$
E\left[Y_{\ell,k}\left|\sum_i X_{i,k}\right|^p\right]\le Q^{p(\ell+1)}\sum_i E|X_{i,k}|^p\le 2^{1+\ell}\sum_i E|X_{i,k}|^p
$$
merely because the absolute value of the sum (if not zero) is less than $Q^{\ell+1}$ times the absolute value of any non-zero term in it.

Step 3: Summing over k with fixed $\ell$. The crucial point here is that if we have any sequence of random variables $F_k$ such that the absolute value of $F_k$ is either $0$ or between $Q^{\ell+k}$ and $Q^{\ell+k+1}$, then
$$
\left|\sum_k F_k\right|^p\le 4^p\max_k|F_k|^p\le 4^p\sum_k|F_k|^p\,.
$$

It follows that
$$
E\left|\sum_k \left[Y_{\ell,k}\sum_i X_{i,k}\right]\right|^p\le 4^p\cdot 2^{1-|\ell|}\sum_k \sum_i E|X_{i,k}|^p= 4^p\cdot 2^{1-|\ell|}\sum_i E|X_{i}|^p\,,
$$
i.e.,
$$
\left\|\sum_k \left[Y_{\ell,k}\sum_i X_{i,k}\right]\right\|_{L^p}\le 4\cdot 2^{(1-|\ell|)/p}\left[\sum_i E|X_{i}|^p\right]^{1/p}\,.
$$

Step 4: Use Minkowski's inequality when summing over $\ell$ now and observe that
$$
\sum_\ell\sum_k \left[Y_{\ell,k}\sum_i X_{i,k}\right]=\sum_i X_i\,.
$$

*Edit:* Chasing the constant.

The technique outlined above is often useful as a replacement of interpolation when the latter fails to work directly for some reason. However, in this case we can utilize the complex interpolation tool in its original form for *symmetric* random variables. Here is how it can be done (I'll even assume that $X_i$ are complex, though it does not matter):

Normalize the variables so that $\sum_i E|X_i|^p=1$. Write each $X_i$ as $e^{\frac 1pG_i}u_i$ with real $G_i$ and $|u_i|=1$ so $\sum_i Ee^{G_i}=1$. Also choose $Y=e^{\frac 1qH}v$ so that $E|Y|^q=Ee^{H}=1$ and
$$
E(Y\sum_i X_i)=\left\|\sum_i X_i\right\|_{L^p}\,.
$$
Now just consider $X_{i,z}=e^{zG_i}u_i$ and $Y_z=e^{(1-z)H}v$ as in the classical proof of the Riesz-Thorin interpolation theorem. The trick is that this modification of $X_i$ does not preserve the mean $0$ property in general but *does* preserve symmetry (which informally merely means that for every given value of $G_i$ we have every value of $u_i$ together with its opposite). Also, since $X_{i,z}$ is constructed based on the values of $X_i$ alone, the pairwise independence is also preserved.

Now we can consider $F(z)=E(Y_z\sum_i X_{i,z})$ as an analytic in the strip $\frac 12\le\Re z\le 1$ function, as usual, and use the trivial endpoint estimates ($p=1$ and $p=2$) to get the whole range with constant $1$ for the *symmetric* case.

The crude reduction to the symmetric case yields then the extra factor $2^p$, but it one thinks for just a moment, one realizes that this constant is actually improving, not deteriorating, as we move from $1$ to $2$. So the question becomes the following:

Let $a_p$ and $A_p$ be the best constants in the inequality
$$
a_pE|Z|^p \le E|Z-Z'|^p\le A_pE|Z|^p
$$
where $Z$ is a mean $0$ random variable (real, complex, or Hilbert-space valued: it should, probably, be irrelevant though the original question was about the real case). What is the ratio $A_p/a_p$? It is easy to see that $a_1=1$ while $A_1=2$ explaining the jump discontinuity at $1+$, but it is also easy to see that $a_p,A_p\to 2$ as $p\to 2$, so there is no discontinuity there. I suspect that those values are just known, so I'll wait a bit before thinking of what I can prove myself here.

*Edit 2*. Two more remarks.

- One can also prove by complex interpolation a slightly more general inequality
$$
E\left|\sum_i(X_i-EX_i)\right|\le 2^{2-p}\sum_iE|X_i|^p
$$
for arbitrary $X_i$ (not necessarily mean $0$). All it takes is to notice that $X_{i,z}-EX_{i,z}$ depends on $z$ analytically as well. This circumvents the reduction to the symmetric case though the symmetric case with sharp constant $1$ may be of interest by itself.

*Edit: Adding some details by Iosif's request:*

For a random variable $X$, define $X_{i,z}=X^{[pz]}-EX^{[pz]}$ (with $X^{[pz]}$ defined as in Iosif's post). Put $S_z=\sum_i X_{i,z}=\sum_i(X_i^{[pz]}-EX_i^{[pz]})$. Then $X_{i,1/p}=X_i-EX_i$ and $S_{1/p}=S=\sum_i(X_i-EX_i)$. Put $Y_z$ to be exactly the same as in Iosif's post and define
$$
F(z)=E \bar Y^{[q(1-z)]}S_z
$$
Then $F(1/p)=\|S\|_p$ while for $\Re z=1$, we have $|F(z)|\le 2$ ($\|Y_z\|_\infty\le 1$ and $\|\sum_i X_{i,z}\|_1\le 2$), while for $\Re z=1/2$, we have $\|Y_z\|_2\le 1$ and $\|S_z\|_2^2=\sum_i\|X_{i,z}\|^2\le\sum_i\|X_i^{[pz]}\|_2^2=1$, so $|F(z)|\le 1$ ($X_{i,z}$ are independent mean zero, hence orthogonal, and subtracting the mean can only reduce the $L^2$-norm (but can raise the $L^1$-norm twice)).

Now consider $F(z)2^{1-2z}$ ($z-1$ in the comment was a typo). This function is bounded by $1$ on both lines, so everywhere. For $z=1/p$, this gives
$$
\|S\|_p=F(1/p)=2^{\frac 2p-1}(F(1/p)2^{1-\frac 2p})\le 2^{\frac 2p-1}
$$
and, raising both sides to the $p$-th power, we get
$$
E|S|^p\le 2^{2-p}\,.,
$$

- Let $\varepsilon>0$. Consider $m$ sets $F_i$ of measure $\epsilon$ with pairvise intersections of measure $\varepsilon^2$ but no triple intersections (they are easy to construct by induction when $m\varepsilon<1$). Then $X_i=\chi_{F_i}-\varepsilon$ are pairwise independent with mean $0$ and $E|X_i|^p=\varepsilon(1-\varepsilon)^p+(1-\varepsilon)\varepsilon^p$. On the other hand,
$$
E\left|\sum_i X_i\right|^p=
\\
m(\varepsilon-(m-1)\varepsilon^2)(1-m\varepsilon)^p+\frac{m(m-1)}2\varepsilon^2(2-m\varepsilon)^p
\\
+\left(1-m\varepsilon+\frac{m(m-1)}2\varepsilon^2\right)(m\varepsilon)^p\,.
$$
Considering the regime $\varepsilon\to 0$, $m\varepsilon\to x\in(0,1)$, we get
$$
(x-x^2)(1-x)^p+\tfrac{x^2}2(2-x)^p+(1-x+\tfrac{x^2}2)x^p\le C_p x\,.
$$
Thus,
$$
C_p\ge (1-x)^{p+1}+\tfrac x2(2-x)^p+(1-x+\tfrac{x^2}2)x^{p-1}
$$
for every $x\in(0,1)$.

The graph of the maximum of the RHS over $x\in(0,1)$ is below (in red) compared to $2^{2-p}$ (in green) but it is clear that for $x$ close to $0$ the RHS tends to a number close to $2$ when $p\to 1$ (the first and the third term will be about $1$ each), so $C_{1+}=2$.