**Question 1.** Suppose that $0^{\#}$ exists. Are there set generic filter $g,h$ over $L$ ($g,h \in V$) such that for any $f$ ($\in V$) which is generic over $L$ we have that $L[g] \cup L[h] \not \subseteq L[f]$?

In other words: Does $0^{\#}$ imply the failure of the upward directedness of the set generic universe over $L$?

I have the following partial result:

**Corollary 2.**  Suppose $0^{\#}$ exists. Let $\kappa$ be the least innaccessible
  cardinal of $L$. Then there are $g,h$ which are $\mathbb{C}$-generic
  over $L$ ($\mathbb{C}$ is Cohen forcing) such
  that for any $f$ which is $\mathbb{P}$-generic over $L$ for some
  $\mathbb{P} \in J_{\kappa}$ we have
  $L[g] \cup L[h] \not \subseteq L[f]$.

That observation came out of:

**Proposition 3.** (Woodin)  Let $M$ be a countable transitive model of $\mathrm{ZFC}$ and let
  $\mathbb{C}$ be Cohen forcing. Then there are $\mathbb{C}$-generic
  filters $g,h$ over $M$ such that for any transitive $\mathrm{ZFC}$ model $N$
  with the same ordinal height of $M$ we have
  $M[g] \cup M[h] \not \subseteq N$.

  In particular there is no set generic filter $f$ over $M$ such that
  $M[g] \cup M[h] \subseteq M[f]$.

Proof.  Let $\alpha = M \cap \mathrm{Ord} < \omega_{1}$ and fix a bijection
  $f \colon \omega \to \alpha$. Define
  $E \subseteq \omega \times \omega$ via
  $$
    (m,n) \in E \iff f(m) \in f(n).
  $$
  Let $z \in ^{\omega}2$ code $E$ in an absolute fashion. E.g. we may let
  $$
    z(k) = 1 \iff k = 2^{m}\cdot3^{n} \wedge (m,n) \in E.
  $$
  Clearly any transitive model having $z$ as an element has $E$ as an
  element as well and, since $E$ is a well-order of order type
  $\alpha$, must also have $\alpha$ as an element.

It now suffices to construct Cohen reals $c,d$ over $M$ such that
  they combined code the real $z$.  We
  construct $c,d \in ^{\omega}2$ as follows:

  Since $M$ is countable, we may fix an enumeration
  $(D_{k} \mid k < \omega)$ of all dense subsets of $\mathbb{C}$ that
  are elements of $M$.  Let $c_{0} \in D_{0}$ and let
  $d_{0} = (0^{\mathrm{length}(c_{0})}) ^{\frown} (1) ^{\frown} z(0) ^{\frown}
  y_{0}$ for some $y_{0}$ such that $d_{0} \in D_{0}$. Given
  $c_{k}, d_{k}$ for some $k < \omega$ we let
  $$
    c_{k+1} = c_{k} ^{\frown} (0^{\mathrm{length}(d_{k})- \mathrm{length}(c_{k})}) ^{\frown} (1) ^{\frown} x_{k+1}
  $$
  for some $x_{k+1}$ such that $c_{k+1} \in D_{k+1}$ and
  $$
    d_{k+1} = d_{k} ^{\frown} (0^{\mathrm{length}(c_{k+1}) - \mathrm{length}(d_{k})}) ^{\frown} (1) ^{\frown} y_{k+1}
  $$
  for some $y_{k+1}$ such that $d_{k+1} \in D_{k+1}$.  We then let
  $c = \bigcup_{k < \omega} c_{k}$ and
  $d = \bigcup_{k < \omega} d_{k}$. The initial segments of $c$ and $d$ look as follows
  $$
    \begin{array}{cc|c|c|c|c}
      c = & c_{0}  & 0 \ldots 0 & 1 \ x_{1} & 0  \ldots 0 & \ldots \\
      d = & 0 \ldots 0 & 1 \ z(0) \ y_{0} & 0 \ldots 0 & 1 \ z(1) \ y_{1} & \ldots
  \end{array}
  $$
  and the blocks of $0$'s in $c$ and $d$ now allow us to reconstruct $z$ from
  $c,d$ via a recursive function. (Q.E.D.)

Proof of Corollary 2.    Since $0^{\#}$ exists, $\kappa$ exists and is countable in
  $V$. Since all dense subsets of $\mathbb{C}$ that $L$ can see are in
  $J_{\omega_{1}^{L}} \subseteq J_{\kappa}$, there are only countably
  many such dense sets. Just like before we fix a real $z$ that codes
  the countable ordinal $\kappa$ and construct $g,h$ which are
  $\mathbb{C}$-generic over $L$ such that they combined code
  $z$. Suppose there were some $\mathbb{P} \in J_{\kappa}$ and some
  $f$ which is $\mathbb{P}$-generic over $L$ with
  $L[g] \cup L[h] \subseteq L[f]$. $g,h,f$ are, of course,
  $\mathbb{C}$(respectively $\mathbb{P}$)-generic over $J_{\kappa}$
  and we would have $g,h \in J_{\kappa}[f]$. But now $J_{\kappa}[f]$
  would have to contain $\kappa$ as an element. Contradiction! (Q.E.D.)