Conditional distribution/independence Suppose $X,Y$ and $Z$ are random elements on $(\Omega,\mathcal{A},\mathit{P})$ taking values in the Borel spaces $U,V$ and $V$ respectively. Moreover, let $\mathcal{F}\subset \mathcal{A}$ be a $\sigma-$field, and assume that regular versions $\mu$, $\nu$ of the conditional distributions $P[Y\in\cdot|\mathcal{F}]$ and $P[Z\in\cdot|\mathcal{F}]$ respectively exist. Now, if $X$ is $\mathcal{F}$ measurable $f:U\times V\to\mathbb{R}$ is measurable and bounded, do the following implications hold?
1.) 
$Y\,\bot_\mathcal{F}Z\hspace{5pt}\Longrightarrow f(X,Y)\,\perp_\mathcal{F}f(X,Z)$ 
where $\perp_\mathcal{F}$ denotes conditional independence with respect to $\mathcal{F}$,
2.)
$Y$ and $Z$ have the same conditional distribution with respect to $\mathcal{F}$ a.s.
$\Longrightarrow$ 
$f(X,Y)$ and $f(X,Z)$ have the same conditional distribution with respect to $\mathcal{F}$ a.s.
My attempt:
2.)
By hypothesis, we have $\mu(A)=\nu(A)$ a.s. for $A\in\mathcal{A}$. This implies 
$$
\int_\Omega f\, d\mu=\int_\Omega f\,d\nu\hspace{5pt}\text{a.s.}
$$
Therefore, by disintegration
$$
\mathit{P}(f(X,Y)\in A|\mathcal{F})=\int I_A(f(X,y))\,\mu(dy)=\int f(X,y)\,\nu(dy)=\mathit{P}(f(X,Z)\in A|\mathcal{F}),
$$
where the equalities hold a.s. This proves 2.) (I think :)).
In order to prove 1.) via disintegration, I require
$$
(\mu\otimes\nu)(C)=\mathit{P}[(Y,Z)\in C|\mathcal{F}]\hspace{5pt}\text{a.s.  for}\hspace{5pt}C\in V\otimes V.
$$
where $\mu\otimes\nu$ is the pointwise product measure and $V\otimes V$ is the product $\sigma$-field. However, I am unsure whether this holds.
I appreciate any feedback.
 A: $\newcommand{\F}{\mathcal{F}}
\newcommand{\G}{\mathcal G}
\newcommand{\HH}{\mathcal H}$
The answer to both your questions is yes, even without assuming the existence of regular versions of conditional distributions. 
Indeed, let $\G$ and $\HH$ be the sigma-algebras over $U$ and $V$, respectively, with respect to which $X$ is a random element of $U$, and $Y$ and $Z$ are random elements of $V$. In what follows, take arbitrarily $A\in\G,A_1\in\G,A_2\in\G,B\in\HH,C\in\HH,F\in\F$. 
Then 
\begin{equation}
 P(X\in A,Y\in B|\F)=E(1_{X\in A,Y\in B}|\F)=E(1_{X\in A}1_{Y\in B}|\F)
 =1_{X\in A}E(1_{Y\in B}|\F)=1_{X\in A}P(Y\in B|\F). \tag{1}
\end{equation}
Similarly, 
\begin{equation}
 P(X\in A,Z\in C|\F)=1_{X\in A}P(Z\in C|\F)  \tag{2}
\end{equation}
and 
\begin{equation}
 P(X\in A,Y\in B,Z\in C|\F)  
 =1_{X\in A}P(Y\in B,Z\in C|\F). \tag{3}
\end{equation}
If $Y$ and $Z$ have the same conditional distributions given $\F$, then $P(Y\in B|\F)=P(Z\in B|\F)$ and hence, by (1) and (2), $P(X\in A,Y\in B|\F)=P(X\in A,Z\in B|\F)$, so that $(X,Y)$ and $(X,Z)$ have the same conditional distributions given $\F$, which implies that $f(X,Y)$ and $f(X,Z)$ have the same conditional distributions given $\F$. This settles your second question. 
Now suppose that $Y\perp_\F Z$, so that 
\begin{equation}
 P(Y\in B,Z\in C|\F)=P(Y\in B|\F)P(Z\in C|\F). \tag{4}
\end{equation}
Then 
\begin{multline*}
 P((X,Y)\in A_1\times B,(X,Z)\in A_2\times C|\F) \\ 
 =P(X\in A_1\cap A_2,Y\in B,Z\in C|\F) \\ 
 =1_{X\in A_1\cap A_2}P(Y\in B,Z\in C|\F) \\ 
 =1_{X\in A_1\cap A_2}P(Y\in B|\F)P(Z\in C|\F) \\ 
 =1_{X\in A_1}P(Y\in B|\F)1_{X\in A_2}P(Z\in C|\F) \\ 
 =P(X\in A_1,Y\in B|\F)P(X\in A_2,Z\in C|\F) \\ 
 =P((X,Y)\in A_1\times B|\F)P((X,Z)\in A_2\times C|\F);   
\end{multline*}
here, the second equality immediately follows from (3), the third equality immediately follows from (4), and the fifth equality immediately follows from (1) and (2). Thus, $(X,Y)\perp_\F(X,Z)$ and hence $f(X,Y)\perp_\F f(X,Z)$. 
This settles your first question as well. 
