Can a 3D joint distribution be reconstructed by 2D marginals? Suppose we know p(x,y), p(x,z) and p(y,z), is it true that the joint distribution p(x,y,z) is identifiable? I.e., there is only one possible p(x,y,z) which has above marginals?
 A: In the same spirit as whuber's answer,
Consider jointly continuous random variables $U, V, W$  with joint density function 
\begin{align}
f_{U,V,W}(u,v,w) = \begin{cases} 2\phi(u)\phi(v)\phi(w)
& ~~~~\text{if}~ u \geq 0, v\geq 0, w \geq 0,\\
& \text{or if}~ u < 0, v < 0, w \geq 0,\\
& \text{or if}~ u < 0, v\geq 0, w < 0,\\
& \text{or if}~ u \geq 0, v< 0, w < 0,\\
& \\
0 & \text{otherwise}
\end{cases}\tag{1}
\end{align}
where $\phi(\cdot)$ denotes the standard normal density function.
It is clear that $U, V$, and $W$ are dependent
random variables. It is also clear that they are not
jointly normal random variables.
However, all three pairs $(U,V), (U,W), (V,W)$
are pairwise independent random variables: in fact,
independent standard normal random variables (and thus
pairwise jointly normal random variables).
In short,
$U,V,W$ are an example of pairwise independent but not
mutually independent standard normal random variables.
See this answer of mine
for more details.
In contrast, if $X,Y,Z$ are mutually independent standard normal random variables, then they are also pairwise independent random variables but their joint density is
$$f_{X,Y,Z}(u,v,w) = \phi(u)\phi(v)\phi(w), ~~u,v,w \in \mathbb R \tag{2}$$ which is not the same as the joint density in $(1)$. So, NO, we cannot deduce the trivariate joint pdf from the bivariate pdfs even in the case when the marginal univariate distributions are standard normal and the random variables are pairwise independent.
A: You're basically asking if CAT reconstruction is possible using only images along the 3 main axes.
It is not... otherwise that's what they would do. :-) See the Radon transform for more literature.
A: No.  Perhaps the simplest counterexample concerns the distribution of three independent $\text{Bernoulli}(1/2)$ variables $X_i$, for which all eight possible outcomes from $(0,0,0)$ through $(1,1,1)$ are equally likely.  This makes all four marginal distributions uniform on $\{(0,0),(0,1),(1,0),(1,1)\}$.
Consider the random variables $(Y_1,Y_2,Y_3)$ which are uniformly distributed on the set $\{(1,0,0),(0,1,0), (0,0,1),(1,1,1)\}$.  These have the same marginals as $(X_1,X_2,X_3)$.

The cover of Douglas Hofstadter's Godel, Escher, Bach hints at the possibilities.

The three orthogonal projections (shadows) of each of these solids onto the coordinate planes are the same, but the solids obviously differ.  Although shadows aren't quite the same thing as marginal distributions, they function in rather a similar way to restrict, but not completely determine, the 3D object that casts them.
