None of them are (generally) true, and this is easy to prove by counterexamples.
If $E[XY]\ne0$,$E[YZ]\ne0$ then can we say
$$E[XZ]\ne0$$
Suppose we have: $$X=A+B$$ $$Y=A+C$$ $$Z=C+D$$
where $A,B,C,D$ are independent random variables with mean 0 (but non-zero finite variance). Then we have $E[XY]=E[A^2]$, $E[YZ]=E[C^2]$ and $E[XZ]=0$.
Or, in words: because the correlations between $X$ and $Y$ on the one hand, and $Y$ and $Z$ on the other, are mediated by different shared variables ($A$ vs $C$), nothing is shared between $X$ and $Z$.
Alternatively $E[XY]=0$,$E[YZ]=0$ then can we say
$$E[XZ]=0$$
This one is even easier. Just suppose $Z=X$, while $Y$ is independent of both. Then clearly the two conditions are satisfied, and $E[XZ]=E[X^2]=E[Z^2]$.
Or even $E[XY]=0$,$E[YZ]\ne0$ then can we say
$$E[XZ]=0$$
Using the same set of random variables as before, suppose:
$$X=A+B$$
$$Y=C+D$$
$$Z=A+D$$
Then $X$ and $Y$ don't share any variance, i.e. $E[XY]=0$, $Y$ and $Z$ share variance through $D$, s.t. $E[YZ]=E[D^2]\neq 0$, and $X$ and $Z$ share variance through $A$, s.t. $E[XZ]=E[A^2]\neq 0$.