Oracle for an inverse function
I am skipping the first question, question A, as it is not clear to me
how it can make sense formally, since I am not familiar with the
formalizations the OP mentions in a comment: Are there established complexity classes with real numbers?

I am only adressing the relation between question A and question B,
proving that question B can be reduced to question A, which the OP considers of interest in itself. And since we are supposed to deal with only one question at a time, this follows the local rules. :)

Question B reduces to question A : B ≤ A

First I will rephrase both questions, so as to add a missing
hypothesis (bijectivity of the function, though existence of an
inverse implies it), and also to use notations that will make the proof
easier to state.

Question A: Let $F$ be a bijective, monotonically increasing real
function from $[0,1]$ onto $[0,1]$. Given an oracle to compute $F$ on
any input and a number $y\in[0,1]$, is there a way to calculate the
inverse function $F^{-1}(y)$ in a finite number of calls to the $F$
oracle?

Question B: Let $H, K$ be two functions like $F$ above, and let $G$ be
their average: $G(x)=[H(x)+K(x)]/2$. Given an oracle to compute
$H^{-1}$ and an oracle to compute $K^{-1}$ and a number $y\in[0,1]$,
is there a way to calculate $G^{-1}(y)$ in a finite number of calls to
the $H$ and $K$ oracles.?

Note: all functions considered are bijective monotonic, but for
readability we leave implicit the related arguments, though they do matter in some of the inferences below.

The idea of the proof is that, if $x=G^{-1}(y)$ is a solution to
problem B, then there are two numbers $h,k\in[0,1]$ such that


$y=(h+k)/2$, and
$h=H(x)$, and $k=K(x)$ for some $x\in [0,1]$


We try to resolve the problem by defining a variable $u\in[0,1]$, so
that $h$ and $k$ are computed from $u$ to take all possible values
while $u$ varies from $0$ to $1$.

From $y=(h+k)/2$ we get $h+k=2y$. So we define: $h_u=2yu$ and
$k_u=2y(1-u)$.
This ensures that $y=(h_u+k_u)/2$, but will let $h_u$ or $k_u$ become greater than $1$ when $y$ is greater than $1/2$.

So, we must actually consider two somewhat similar cases, depending on the value of $y$, after noting that $h_u$ and $k_u$ are both monotonically increasing functions of $u$.

case 1: $y\leq 1/2$

What we want is to find $h$ and $k$ such that $h=H(x)$ and $k=K(x)$
for some value $x$.

We define the function
$D(u)=H^{-1}(h_u)-K^{-1}(k_u)=H^{-1}(2yu)-K^{-1}(2y(1-u))$
Note that $H^{-1}(0)=K^{-1}(0)=0$.
Hence $D(0)=H^{-1}(0)-K^{-1}(2y)=-K^{-1}(2y)$,
and $D(1)=H^{-1}(2y)-K^{-1}(0)=H^{-1}(2y)$.

So the function $D(u)$ is increasing onto $[-K^{-1}(2y),H^{-1}(2y)]$,
and the solution to finding the value of $u$ that corresponds to the
right values of $h$ and $k$ is given by the equation $D(u)=0$.

However, we intend to resolve it using an algorithm to solve
question A, so we must normalize the interval to make it $[0,1]$.

For that we define
$F(u)=\frac{D(u)+K^{-1}(2y)}{H^{-1}(2y)+K^{-1}(2y)}$
The function $F$ satisfies the conditions of question A,
and we remark that $D(u)=0$ iff
$F(u)=\frac{K^{-1}(2y)}{H^{-1}(2y)+K^{-1}(2y)}$.
So we have to compute
$u_0=F^{-1}(\frac{K^{-1}(2y)}{H^{-1}(2y)+K^{-1}(2y)})$, which can be
done using the algorithm answering question A.
Then $G^{-1}(y)=H^{-1}(2yu_0)$, or to summarize in one formula:

$$G^{-1}(y)=H^{-1}(2yF^{-1}(\frac{K^{-1}(2y)}{H^{-1}(2y)+K^{-1}(2y)}))$$

Each computation of $F(u)$ uses only a finite (small) number of calls to the
oracles for $H^{-1}$ and $K^{-1}$. Thus a solution to question B can
be computed with a finite number of oracle calls if that is the case
for problem A.

case 2: $y&gt;1/2$

This case is a bit more complicated, as different bounds must be found
for varying $u$ so that $h$ and $k$ do not get greater than $1$, since
the functions $H$ and $K$ are defined only onto $[0,1]$.

Since $y=(h_u+k_u)/2$, we have $h_u=2y-k_u$. Thus, $k_u\leq 1$ implies
$h_u\geq 2y-1$.

The same reasonning holds, exchanging $h$ and $k$, so that we have
$h_u,k_u\in [2y-1, 1]$. Since $u=h_u/(2y)$, we have $u\in[u_0,u_1]$
with $u_0=1-1/(2y)$ and $u_1=1/(2y)$.

In order to have a variable $v$ varying in $[0,1]$, we normalize by
defining $v=(u-u_0)/(u_1-u_0)$. But
$u_1-u_0=1/(2y)-(1-1/(2y))=(1-y)/y$
which gives $v=(u-u_0)y/(1-y)$.
Hence, with $u=v(1-y)/y+u_0$, $u$ is
a function of $v$ monotonically increasing onto $[u_0,u_1]$ when
$v\in[0,1]$.

Thus, considering $h$ and $k$ as functions of $v$ we have
$h_v=2y(v(1-y)/y+u_0)$ and $k_v=2y(1-(v(1-y)/y+u_0))$, with $u_0=1-1/(2y)$

Simplifying we get $h_v=2y-1+2v(1-y)$ and $k_v=1-2v(1-y)$, for $v\in[0,1]$

From here on the reasonning is the same as for the first case, since monotonically increasing functions are closed under composition, the variable being $v$ rather than $u$, considering the function:
$D(v)=H^{-1}(h_v)-K^{-1}(k_v)=H^{-1}(2y-1+2v(1-y))-K^{-1}(1-2v(1-y))$  

I hope that there is no error in symbol manipulations, but the line of reasonning should not be affected if that were the case.

The proof can be simplified by using a lemma that generalizes the question A, which is here a hypothesis, to any strictly increasing function from any closed interval to any other closed interval.