Even with the comment this is still not entirely clear. The only way I can make sense of the question is as follows:
Let $\Phi$ be a root system and $\lbrace \alpha_1, ..., \alpha_n \rbrace=\Delta \subset \Phi$ (note $n = rank(\Phi)$) a chosen set of simple roots. Let
$$\langle \alpha_i^\vee,\alpha_j\rangle\langle \alpha_i,\alpha_j^\vee\rangle\alpha_i-3\alpha_i=\delta_{ij}\alpha_j$$
for all $\alpha_i, \alpha_j \in \Delta$. Then what can we say about $\Phi$?
The answer is that $\Phi$ must be the root system of type $A_1$ (with $n=1$) or of type $G_2$ (with $n=2$), and no other. Namely, for $n=1$ there is only that root system (and with your reasoning it satisfies the criterion trivially); as soon as $n\ge 2$, with your reasoning we see that e.g. the angle between the roots $\alpha_1, \alpha_2$ is $5\pi/6$ or $\pi/6$; it must be the first option because otherwise they could not both be simple roots; so they span a root system of type $G_2$. Now every root system is a direct sum of irreducible ones; and by inspection, none of the other irreducible ones contains a subsystem of type $G_2$. This means that if the rank of $\Phi$ is $\ge 3$, it must be of the form $G_2 \oplus \Phi'$ with some other root system $\Phi'$ of rank $n-2$. However, then any of the roots $\alpha_3, ..., \alpha_n$ are in $\Phi'$, hence orthogonal to both $\alpha_1, \alpha_2$, in particular the above criterion is not satisfied e.g. for $i=1, j=3$ where $\langle \alpha_i^\vee,\alpha_j\rangle\langle \alpha_i,\alpha_j^\vee\rangle =0$, contradiction.
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
You may look at the book of Banyaga on the question. He has shown that the connected component of the group of diffeomorpisms of a compact manifold is simple (or more generally diffeomorphisms with compact support), so its Lie algebra is simple.
https://www.springer.com/gp/book/9780792344759