Here is an answer to question (2), strongly inspired by Dave Benson's comment:
Theorem Let $A$ be any ring and let $M$ be a finite length $A$-module. Then the lattice of $A$-submodules is distributive iff $M$ does not have a subquotient of the form $S^2$, for $S$ simple.
Proof First, suppose that $M$ has an $S^2$ subquotient, say $X \subset Y \subseteq M$ with $Y/X \cong S^2$. Then the interval $[X,Y]$ in the submodule lattice of $M$ is isomorphic to the submodule lattice of $S^2$. In particular, the submodule lattice of $S^2$ contains $S \oplus \boldsymbol{0}$, $\boldsymbol{0} \oplus S$ and $\Delta:= \{ (s,s) : s \in S \}$. Then $\{ \boldsymbol{0}, S \oplus \boldsymbol{0}, \boldsymbol{0} \oplus S, \Delta, S^2 \}$ form a copy of the $M_3$ diamond lattice, which is not distributive.
Conversely, suppose that the submodule lattice of $M$ is not distributive. Then it must contain either a copy of $M_3$ or $N_5$. But the submodule lattice is modular, so it cannot contain $N_5$, so it must contain $M_3$. In other words, there are submodules $X \subset W_1, W_2, W_3 \subset Y$ with $W_i \cap W_j = X$ and $W_i + W_j = Y$ for all $i \neq j$. Put $Z = Y/X$ and $V_i = W_i/X$. So $Z$ has submodules $V_i$ such that $W_i \cap W_j = \{ 0 \}$ and $W_i + W_j = Y/X$ for $i \neq j$. Then we have $$V_2 \cong Z/V_1 \cong V_3 \cong Z/V_2 \cong V_1 \cong Z/V_3$$ so $V_1 \cong V_2 \cong V_3$ and $Z \cong V_1^2$. Let $S$ be a simple quotient of $V$, then $S^2$ is a quotient of $Z$ and is a subquotient of $Y/X$. $\square$
Corollary: The submodule lattice of $M_1 \oplus M_2 \oplus \cdots \oplus M_r$ is distributive iff (1) the submodule lattice of each $M_i$ is individually distributive and (2) for $i \neq j$, the modules $M_i$ and $M_j$ have no common simple subquotient.
Now, in the preprojective Dynkin case, every simple module is a subquotient of every indecomposable projective module. So this simplifies to
Corollary: In the original Dynkin case that the OP asked about, if a projective module has distributive submodule lattice then it is indecomposable.
I have some thoughts about the other parts, but I need to work on other things for now.