Took some work and some scouring through the literature, but we got there in the end. Both questions are answered positively — though my brain still needs some time to digest the answer for question one, and make sure that it's really really true, so, eh, approach with care.
First question:
Whenever I say parabolic here, I mean not the Knapp definition, but the one in terms of complexification, see the OP.
From Lemma in Section 3.2 from Wolf, Koranyi, we can extract the following (heavily paraphrased, but hopefully equivalent):
Let $\mathfrak{g}$ be a real semisimple Lie algebra and $\mathfrak{q}$ a parabolic subalgebra. Then there is some Cartan decomposition
$$\mathfrak{g} = \mathfrak{t} \oplus \mathfrak{p} $$
some maximally noncompact ("maximally split") Cartan subalgebra $\mathfrak{h} = \mathfrak{t} \oplus \mathfrak{a}$ of $\mathfrak{g}$, where
$$\mathfrak{t} \subset \mathfrak{k} \text{ (the "totally nonsplit" part)}, \quad \mathfrak{a} \subset \mathfrak{p} \text{ (the "totally split" part) },$$
a subspace $\mathfrak{a}' \subset \mathfrak{a}$ and a choice of positive roots $P$ in the restricted root space decomposition of $(\mathfrak{g}, \mathfrak{a}')$ so that
$$\mathfrak{q} = \mathfrak{g}_0 \oplus \bigoplus_{\alpha \in P} \mathfrak{g}_\alpha, \quad
\mathfrak{g}_\alpha = \{x \in \mathfrak{g} : [a,x] = \alpha(a) \cdot x \quad \forall a \in \mathfrak{a}'\}.
$$
(Careful: In the source, the root space decomposition is carried out in the complexification $\mathfrak{g}_\mathbb{C}$, but we can also carry it out in the real setting, since the ad-action of elements in $\mathfrak{a}' \subset \mathfrak{a}$ is real diagonalizable. This is always necessary for the restricted root space decomposition.)
Very verbose, but in the end, in the above notation, every parabolic subalgebra contains $Z_\mathfrak{k}(\mathfrak{a}) \oplus \mathfrak{a}$ in the $\mathfrak{g}_0$-component and some choice of positive restricted roots of $\mathfrak{a}$ in the $\bigoplus_{\alpha \in P} \mathfrak{g}_\alpha$-component. Hence every parabolic subalgebra contains some subalgebra of the form $\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$. And indeed, the subalgebras $\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$ are parabolic, since their complexification contains a Borel algebra associated to the complexification of the Cartan subalgebra $\mathfrak{a} \oplus \mathfrak{t}$.
Hence, Knapp's minimal parabolic subalgebras are exactly the minimal parabolic subalgebras in the usual sense.
Second question:
In Bourbaki, Chapter VIII, Exercise 3a for §5, we learn:
If $\mathfrak{q}, \mathfrak{p}$ are parabolic subalgebras of a semisimple (real or complex) Lie algebra $\mathfrak{g}$, and $\mathfrak{q} \subset \mathfrak{p}$, then the radical of $\mathfrak{p}$ is contained in the radical of $\mathfrak{q}$.
And in Bourbaki, Chapter VIII, §10, Corollary 2, we learn:
Every subalgebra $\mathfrak{n}$ of a (real or complex) Lie algebra $\mathfrak{g}$, consisting only of nilpotent elements of $\mathfrak{g}$, is contained in the nilradical of a parabolic subalgebra $\mathfrak{q}$.
As a corollary of the two: Given a subalgebra $\mathfrak{n} \subset \mathfrak{g}$, contained in the radical of some parabolic subalgebra $\mathfrak{q}$. But then there is some minimal parabolic $\mathfrak{q}_0 \subset \mathfrak{q}$ with $\mathfrak{n} \subset \text{rad}(\mathfrak{q}) \subset \text{rad}(\mathfrak{q}_0)$.
Wolf, J. A.; Koranyi, A., Generalized Cayley transformation of bounded symmetric domains, Am. J. Math. 87, 899-939 (1965). ZBL0137.27403.
Bourbaki, Nicolas, Elements of mathematics. Lie groups and Lie algebras. Chapters 7 and 8, Berlin: Springer (ISBN 3-540-33939-6). 271 p. (2006). ZBL1181.17001.
Best Answer
In fact, if $L$ is any Lie algebra over any field $k$, and $L_1, L_2$ are subalgebras such that $L = L_1 \oplus L_2$ as $k$-vector spaces, then the $k$-linear map $f$ given by
$$U(L_1) \otimes_k U(L_2) \rightarrow U(L)$$ $$x \otimes y \mapsto xy$$
is an isomorphism of $k$-vector spaces (and hence of $(U(L_1),U(L_2))$-bimodules).
As noticed in the first bullet point in this question, this follows from the Poincaré-Birkhoff-Witt theorem, compare e.g. Bourbaki, Lie Groups and Algebras ch. 1 §7. Indeed, it is a corollary of PBW (in Bourbaki, Corollary 3) that if $(e_1, ..., e_n)$ is an ordered $k$-basis of a Lie algebra $\mathfrak g$, then the monomials $e_1^{r_1} \cdot ... \cdot e_n^{r_n}$ are a $k$-basis of $U(\mathfrak g)$. From this the assertion (which is stated in greater generality in Bourbaki's Corollary 6) follows. Namely, if now $(x_1, ..., x_m)$ is an ordered $k$-basis of $L_1$, and $y_1, ..., y_n$ one of $L_2$, then on the one hand PBW and tensor products over fields tell us that the set of all $x_1^{r_1} \cdot ... \cdot x_m^{r_m} \otimes y_1^{l_1} \cdot ... \cdot y_n^{l_n}$ is a $k$-basis of $U(L_1) \otimes U(L_2)$, on the other hand the images of these elements under $f$, namely, all $x_1^{r_1} \cdot ... \cdot x_m^{r_m} \cdot y_1^{l_1} \cdot ... \cdot y_n^{l_n}$, again by PBW are a $k$-basis of $U(L)$, because by assumption $(x_1, ..., x_m, y_1, ..., y_m)$ is an ordered $k$-basis of $L$.
Now apply this to $L=\mathfrak g, L_1 = \mathfrak{n}^-, L_2= \mathfrak b$.
To see where your supposed counterargument goes wrong (copied from comments and discussion):
Your computation of $xy=0$ for certain $x, y \neq 0$ seems to be happening in $M_3(\mathfrak k)$ (or $End_{\mathfrak k} ({\mathfrak k}^3)$). That matrix ring is an enveloping associative algebra of $\mathfrak{sl}_3$, and maybe likewise would be certain associative subalgebras $B:= \pmatrix{*&*&*\\0&*&*\\0&0&*}$, $N^-:=\pmatrix{0&0&0\\*&0&0\\*&*&0}$ corresponding to $\mathfrak b, \mathfrak n^-$. These algebras are quotients of the respective universal enveloping algebras (by the latter's universal property, although surjectivity might be more subtle), but of course a product being zero in a quotient tells us little about the product being zero in the original ring.
What this computation shows, then, is that even though there are canonical surjections of associative algebras $$U(\mathfrak n^-) \twoheadrightarrow N^-, U(\mathfrak b) \twoheadrightarrow B, U(\mathfrak g) \twoheadrightarrow M_3(k)$$ the map $$N^- \otimes_{\mathfrak k} B \rightarrow M_3(k)$$ induced via those projections from the given map $U(n^-) \otimes U(b) \simeq U(g)$, is no longer an isomorphism. But viewed this way, that is maybe not as surprising.
Full disclosure, I have made the mistake of mixing up standard enveloping matrix algebras with the universal enveloping algebra before, and I still sometimes do calculations in those matrix rings to get a first feeling for the universal enveloping one, but it's important to remember that the matrix algebras (like the above $M_3(\mathfrak k), B, N^-$) are just quotients -- and for that matter, with a lot factored out -- of the "much, much bigger" universal enveloping algebras. In particular, note that the above PBW corollaries also imply that $x^my^n \neq 0$ in $U(\mathfrak g)$ for all $m,n \in \mathbb N$ and any $x \neq 0 \neq y \in \mathfrak g$.