I'm not aware of any generalization that is strong enough to compute the cohomology of such bundles completely, but you can at least use the Borel--Weil--Bott theorem to get some vanishing results. This was already done by Bott in his 1957 Annals paper to show that the cohomology of the tangent bundle $T$ of $G/P$ vanishes in degree > 0. The first step of this computation is to note that the action of $P$ on the tangent space at $P/P$ can be identified, as a $P$-module, with $\mathfrak g/ \mathfrak p$ (with $P$ acting via Ad). Consequently, the short exact sequence of $P$-modules
$$ 0 \to \mathfrak p \to \mathfrak g \to \mathfrak g / \mathfrak p \to 0 $$
yields a short exact sequence of $G$-equivariant vector bundles
$$ 0 \to G \times_P \ \mathfrak p \to G \times_P \ \mathfrak g \to T \to 0 $$
on $G/P$. The middle bundle is trivial, and so we get an isomorphism $H^q(G/P, T) \cong H^{q+1}(G/P, G \times_P \ \mathfrak p)$ for all $q>0$.
It remains to compute $H^\ast(G/P, G \times_P \ \mathfrak p)$. For this, a vanishing result of the Borel--Weil--Bott type is useful. This result states that if you have a $G$-equivariant vector bundle $\mathcal V = G \times_P V$ on $G/P$ then $H^q(G/P, \mathcal V) = 0$ unless $q$ is equal to the index of some nonsingular $\mu + \rho$, where $\mu$ is one of the highest weights of $V$.
For the details, you can either try Bott's original paper, or checkout section 11 of this paper of Dennis Snow. Another paper of potential interest is
P.A. Griffiths, Some geometric and analytic properties of homogeneous complex manifolds. I. Sheaves and cohomology, Acta Math. 110 (1963), 115–155.
It is true that we cannot use an arbitrary property$^1$ $P$ to define a set, in the sense that the collection of all things with property $P$ need not be a set. However, the axiom (scheme) of separation says that we can use an arbitrary property to define a subset: whenever $X$ is a set, the collection $\{x\in X: P(x)\}$ is also a set.
So just take $X=\mathbb{R}$ and $P(x)$ = "There is a Hilbert space such that …". Per Separation, we get that your collection of reals $A$ is in fact a set. And we may now take its supremum.
Note that this illustrates an important point about how $\mathsf{ZFC}$ (and its variants) get around Russell's paradox:
It's size,$^{2}$ not complexity of definition, which controls whether or not a collection is a set or a proper class in $\mathsf{ZFC}$.$^{3}$
Part of the success of $\mathsf{ZFC}$ is due to the ease with which we can in fact verify that something is a set. The only time you'll run into trouble is when you want to form a set which isn't a priori part of some bigger thing you already know is a set; here we may have to think a bit (although the axiom (scheme) of replacement similarly makes things usually very easy, once it's mastered).
EDIT: Per the comments below, let me sketch how to define "complete metric space" in the language of set theory. As you'll see, even the sketch is quite lengthy; if there's a particular point you'd like further information on, I suggest asking a separate question at MSE.
Here's the sequence of definitions we need to whip up:
We need to talk about ordered pairs, functions, and Cartesian products.
We need to build $\mathbb{N}$, so that we can build $\mathbb{Q}_{\ge 0}$, so that we can build $\mathbb{R}_{\ge 0}$; along the way we'll need the notions of equivalence relation and equivalence class, of course.
While the previous two points will be enough to define metric spaces ("An ordered pair $(X,\delta)$ where $X$ is a set and $\delta:X^2\rightarrow\mathbb{R}$ such that [stuff]"), to define complete metric spaces we'll also need the notions of infinite sequence and equivalence relation/class.
The first bulletpoint is standard set-theoretic fare which you'll see treated in the beginning of any text on set theory, so I'll skip it; if you're interested, though, you can start with the wiki page on ordered pairs.
The third is really the first in disguise: an infinite sequence is just a function with domain $\mathbb{N}$.
So all the "meat" is in bulletpoint 2. We proceed as follows:
First, we'll use the von Neumann approach to $\mathbb{N}$: an ordinal is a hereditarily transitive set, ordinals are ordered by $\in$, and the finite ordinals are the ordinals which do not contain any (nonempty) limit ordinal. We then identify $\mathbb{N}$ with the finite ordinals — more jargonily, $\mathbb{N}=\omega$. We define addition and multiplication of ordinals via transfinite recursion as usual.
Next, we consider the equivalence relation $\sim$ on $\omega\times(\omega\setminus\{0\})$ as follows: $$\langle a,b\rangle\sim\langle c,d\rangle \iff ad=bc,$$ and we let $\mathbb{Q}_{\ge0}$ be the set of $\sim$-classes. We lift the ordering on $\omega$ to $\mathbb{Q}_{\ge 0}$ in the obvious way.
Now we're ready to define $\mathbb{R}_{\ge 0}$, via Dedekind cuts: an element of $\mathbb{R}_{\ge 0}$ is a nonempty, downwards-closed, bounded-above subset of $\mathbb{Q}_{\ge0}$. The ordering on $\mathbb{R}_{\ge 0}$ is just $\subseteq$.
With all this in hand, the naive definitions of metric space, Cauchy sequence, and complete metric space translate into the language of set theory directly (if tediously). The point is that all of this is first-order in set theory, with axioms like Powerset (which, despite what they mean intuitively, are indeed first-order) doing the heavy lifting needed to show that the objects we want actually exist at all. (For a bit more about the nuance of "first-order in set theory," see this recent answer of mine.)
$^1$Really I mean "first-order formula," but I don't want to get too much into the details.
$^{2}$Specifically, in a precise sense we have: a class is a proper class iff it surjects onto the class of ordinals. This is not the same as the principle of limitation of size, but it's of similar flavor.
$^3$I should observe that this isn't the only possible response to the need to distinguish between sets and proper classes: there are other set theories (e.g. $\mathsf{NF}$, $\mathsf{GPK^+_\infty}$, ...) which take the other approach. However, these theories make it harder to check whether something is in fact a set.
Best Answer
I don't know about Q1 off the top of my head, but I think that for Q2 you are unlikely to get a good answer. Of course I may have a different view from you as to what "mild" adjectives are reasonable to impose.
The reason I say this is that in your definition of completely reducible you are requiring the summands of the decomposition of $\pi$ to be orthogonal, yet you are allowing $\pi$ to be non-unitary. Therefore I can always take a reducible unitary representation $\sigma: G \to {\mathcal U}(H)$ and then conjugate it with a non-unitary invertible operator on $H$ which will almost surely mess up the the orthogonality inside the decomposition of $\sigma$.
For a concrete example: let $G=\{\pm 1\}$, fix $\varepsilon>0$, and let $\pi$ be the representation of $G$ on ${\bf C}^2$ which sends $-1$ to $\begin{pmatrix} 1 & \varepsilon \\ 0 & -1 \end{pmatrix}$. Since the eigenvectors of this matrix are not orthogonal to each other, $\pi$ does not decompose into orthogonal summands.
Another example showing that "nice" groups can fail to have the property you seek: fix $\varepsilon>0$ and let $\pi: {\bf Z} \to {\rm GL}_2({\bf C})$ be the representation which sends $1$ to $\begin{pmatrix} 1 & \varepsilon \\ 0 & 1 \end{pmatrix}$. This representation is reducible but not decomposable; however, I am not sure if you are ruling out such examples with one of your unspecified "mild" adjectives.
By the way, if there are no examples other than the trivial representation, which satisfy the conditions in Q2 then your Q1 will have a positive answer. This is why I think Q1 is actually not such a good question unless you pin down what your intended definition of "completely reducible" should be.