Does the inverse of the distribution of a modal operator also hold? Does it hold that we have $\Box(\phi\land\psi)\leftrightarrow(\Box\phi\land\Box\psi)$ and does this generalize to arbitrary long conjunctions for some fixed length?
 A: We need to be a little cautious about talking about "modal logic" in general. There are many different modal logics, which are governed by different axioms.
In any modal logic where the necessitation axiom and K axiom are valid (which is true for the most common modal logics), it does indeed hold that $ \square (\phi \wedge \psi) \leftrightarrow (\square \phi \wedge \square \psi)$. This can be verified semantically, but can also be done axiomatically. Below is a sketch of the proof:
$$
\begin{align}
\phi \to (\psi \to (\phi \wedge \psi)) && \text{(PL tautology)} \\
\square (\phi \to (\psi \to (\phi \wedge \psi))) && \text{(necessitation)} \\
\square \phi \to (\square \psi \to \square (\phi \wedge \psi)) && \text{(axiom K)} \\
(\square \phi \wedge \square \psi) \to \square (\phi \wedge \psi)) && \text{(absorption)} \\
\end{align}
$$
$$
\begin{align}
(\phi \wedge \psi) \to \phi  && \text{(PL tautology)} \\
\square ((\phi \wedge \psi) \to \phi)  && \text{(necessitation)} \\
\square (\phi \wedge \psi) \to \square \phi && \text{(axiom K)} \\
(\phi \wedge \psi) \to \psi && \text{(PL tautology)} \\
\square ((\phi \wedge \psi) \to \psi)  && \text{(necessitation)} \\
\square (\phi \wedge \psi) \to \square \psi && \text{(axiom K)} \\
\square (\phi \wedge \psi) \to (\square \phi \wedge \square \psi) 
\end{align}
$$
Once we have established $ \square (\phi \wedge \psi) \leftrightarrow (\square \phi \wedge \square \psi)$, we can show inductively that $\square$ distributes over $\wedge$ for any finite conjunction.
Let our inductive hypothesis be that $\square (\phi_{1} \wedge \cdots \wedge \phi_{n}) \leftrightarrow  (\square \phi_{1} \wedge \cdots \wedge \square \phi_{n})$ is a tautology in modal logic. Then:
$$
\begin{align}
\square (\phi_{1} \wedge \cdots \wedge \phi_{n} \wedge \phi_{n+1}) & \leftrightarrow \square ((\phi_{1} \wedge \cdots \wedge \phi_{n}) \wedge \phi_{n+1}) \\
\square (\phi_{1} \wedge \cdots \wedge \phi_{n} \wedge \phi_{n+1}) & \leftrightarrow ( \square (\phi_{1} \wedge \cdots \wedge \phi_{n}) \wedge \square  \phi_{n+1}) && \text{(by proof above)} \\
\square (\phi_{1} \wedge \cdots \wedge \phi_{n} \wedge \phi_{n+1}) & \leftrightarrow  \square \phi_{1} \wedge \cdots \wedge \square \phi_{n} \wedge \square  \phi_{n+1} && \text{(by IH)}
\end{align}
$$
So the claim holds for any conjunction of finite length $n$.
Something to bear in mind: this doesn't hold for $\vee$!
