Q1: For any $C,D\in Fun(BG,Cat_\infty)$, $Fun(C,D)$ acquires a $G$-action too. Informally, this is described as $F\mapsto gF(g^{-1}-)$, and this is in fact an accurate description if $G$ is a discrete group and $C,D$ are $1$-categories; but more generally, formally you can see it as an internal hom in $Fun(BG,Cat_\infty)$.
Indeed, on $Fun(BG,Cat_\infty)$, $C\times-$ preserves arbitrary colimits (as these are computed pointwise, and it does in $Cat_\infty$), so that it admits a right adjoint, which we can denote by $Fun(C,-)$. Indeed, one may check (by abstract nonsense) that the underlying object of this $\infty$-category with $G$-action is $Fun(C,D)$.
In particular, taking the $G$-fixed points of $Fun(C,D)$ makes sense; but now we really want the homotopy fixed points even if $G$ is discrete and $C,D$ are $1$-categories: you don't want $gF(g^{-1}-) = F$, but you want the data of an isomorphism $\rho_g: F\to gF(g^{-1}-)$ so that the various composites are compatible (in a homotopy coherent fashion)
These homotopy fixed points are $Fun^G(C,D)$. In your situation, $G=B\mathbb Z$ and $C=\Lambda_\infty$ with its $B\mathbb Z$-action.
In particular, note that it is note quite a subcategory, although it does come with a "forgetful" functor $Fun^{B\mathbb Z}(\Lambda_\infty,D)\to Fun(\Lambda_\infty, D)$
Q2: The authors claim that $\mathrm{colim}: Fun(\Lambda_\infty,C)\to C$ is equivariant, which gives their construction by taking $B\mathbb Z$-fixed points.
To see why this functor is actually equivariant takes some work as far as I can tell (although maybe there are simpler solutions) : start by noting that the right adjoint, given by the "diagonal" $C\to Fun(\Lambda_\infty,C)$ is itself equivariant. This is clear from the definition of $Fun(\Lambda_\infty,C)$, as this is an internal hom, so it suffices to check that the projection $C\times \Lambda_\infty\to C$ is equivariant but this is obvious.
So we have a left adjoint whose right adjoint is equivariant. Now the rest of this works in this generality so let's just write it that way : we have an adjunction $L\dashv R$, $L:D\to E$ between $\infty$-categories with $G$-action, where $R:E\to D$ can lives in $Fun(BG,Cat_\infty)$ (and $L$ only in $Cat_\infty$ a priori).
But now one may consider $Adj_R$, the $\infty$-category of $\infty$-categories and right adjoints between them. It follows easily that $R: E\to D$ can be seen as an arrow in $Fun(BG,Adj_R)$. Now $Adj_R\simeq Adj_L^{op}$ in the obvious manner (I write "obvious" but in fact this requires some work to set up $\infty$-categorically), so that we can see $L: D\to E$ as an arrow in $Fun((BG)^{op},Adj_L)$. Composing with the canonical equivalence $(BG)^{op}\simeq BG$, we see that $L: D\to E$ canonically acquires a $G$-equivariant structure.
There's some detail I'm slipping under the rug here: that $D,E$ have the correct $G$-action when you pass from $Adj_R$ to $Adj_L$ and then from $(BG)^{op}$ to $BG$. This follows essentially from the fact that $g$ and $g^{-1}$ act as inverses, and hence are adjoint to one another, essentially uniquely (given that the unit and co-unit are determined).
Again, there may be a simpler way to see that $\mathrm{colim}$ has an equivariant structure, but I'm not sure what that would be (there are certainly other ways to phrase what I wrote, and some of them might actually be simpler)
Q3: $Fun^G(pt,C)$ is $(Fun(pt, C))^{hG}$, but $Fun(pt,C) \simeq C$. One may now check that if $C$ has a trivial $G$-action, then so does $Fun(pt,C)$, and the equivalence above is a trivial-action equivalence. It then follows that in this case, $Fun^G(pt,C) = C^{hG}$.
But now, whenever $C$ is an $\infty$-category with trivial $G$-action, its homotopy fixed points are just $Fun(BG,C)$, so in the case $G=B\mathbb Z$, you get $Fun(BB\mathbb Z,C)= C^{BB\mathbb Z}$.
