I thought this would be a full solution (i.e. showing (1) is $\sqrt{n}/2$), but in fact it only yields (1) is less than or equal to $\sqrt{n}/\sqrt{2}$. I'll still post it after all of the effort, while hoping that someone can improve it!
I believe that a simple change in the Chung-Furedi-Graham-Seymour paper which proves $\sqrt{n}$ is tight can be used to show that your (1) is at most $\sqrt{n}/\sqrt{2}$. Their construction works for all $n$ and uses floor functions, for simplicity I'm going to present a construction only for the case when $n$ is twice the square of an even integer.
Suppose that $n = 2k^2$, $k$ is even, and define a collection of sets $S \subset \mathcal{P}([n])$ as follows. First define $T$ to be the set of all $A \subset [n] = \{1, \ldots, n\}$ for which, for some $0 \leq i < k$, $S$ contains all of $2ki+1, \ldots, 2ki+k$ and none of $2ki+k+1, \ldots, 2ki+2k$. Then $S$ consists of all $B \subset [n]$ for which either
$\bullet$ $|B|$ is even and $B \in T$ or
$\bullet$ $|B|$ is odd and $B \notin T$
For instance, when $k = 2$ and $n = 8$, $T$ contains all sets which either (contain both of $1,2$ and omit both of $3,4$) or (contain both of $5,6$ and omit both of $7,8$), and $S$ consists of all even sets in $T$ and all odd sets in $T^c$.
Claim 1: $|S| = 2^{n-1} \pm 1$. I don't think I should try to write a full proof here, but it's essentially the same proof as for the Chung-Furedi-Graham-Seymour example. For context, their example is the same, but where the definition of $T$ is simpler, instead of containing all of $2ki+1, \ldots, 2ki+k$ and missing all of $2ki+k+1, \ldots, 2ki+2k$ for some $i$, they just require containing all of $2ki+1, \ldots, 2ki+2k$ for some $i$.
They find $|S|$ by inclusion-exclusion, and the key is that for any nonempty proper $\{i_1, \ldots, i_j\} \subsetneq \{0, \ldots, 2k-1\}$, the number of even/odd sets $B$ containing all of $2ki_m + 1, \ldots, 2ki_m + 2k$ for
$1 \leq m \leq j$ is $2^{n-2kj-1}$. That fact is still true in our setting (specifying some elements to be out of $B$ instead of in $B$ has the same effect probabilistically). The only subtlety is what happens when the condition holds for all $i$. In their paper, this forces $B$ to be all of $[n]$ (so even in our setting), and the same condition for us forces $B$ to be exactly the set $\{1, \ldots, k, 2k+1, \ldots, 3k, \ldots, (2k-2)k+1, \ldots, (2k-1)k\}$, of size $k^2$, again even since we assumed $k$ even.
So see their paper for more details, but this should still be true (and I checked for some small $k$).
Claim 2: every set in the induced subgraph for $S$ has no more than $k = \sqrt{n}/\sqrt{2}$ incoming edges and no more than $k = \sqrt{n}/\sqrt{2}$ outgoing edges, and the same is true of the induced subgraph for $S^c$.
Proof: There are eight cases, all similar. Consider an even set $B$ in $S$ with an outgoing edge in $S$, meaning $m \notin B$ so that $B \cup \{m\} \in S$. Then by definition, $B \in T$ and $B \cup \{m\} \notin T$. The only possibility is that $B$ contains all of $2ki+1, \ldots, 2ki+k$ and none of $2ki+k+1, \ldots, 2ki+2k$ for exactly one $i$, and $m \in \{2ki + k + 1, \ldots, 2ki + 2k\}$, so there are exactly $k$ choices for $m$.
If instead there is an incoming edge, then $m \in B$, $B \in T$, and $B - \{m\} \notin T$. Again there is only one possibility; $B$ contains all of $2ki+1, \ldots, 2ki+k$ and none of $2ki+k+1, \ldots, 2ki+2k$ for exactly one $i$, and $m \in \{2ki+1 + 1, \ldots, 2ki + k\}$. Again there are exactly $k$ choices for $m$.
Now suppose instead that $B$ was odd. For an incoming edge, $B \notin T$ and $B \cup \{m\} \in T$. The only possibility is that $B$ nearly satisfied the condition for being in $T$ in that there exists $0 \leq i < k$ for which all but one of $2ki + 1, \ldots, 2ki+k$ are in $B$ and none of $2ki + k+1, \ldots, 2ki+2k$ are in $B$. Each such $i$ yields a single possible value of $m$ (the single missing element), and there are at most $k$ possible values of $i$, so there are at most $k$ possibilities for $m$.
For an outgoing edge, it's the same as above except that now $B$ must contain all of $2ki+1, \ldots, 2ki+k$ and exactly one of $2ki+k+1, \ldots, 2ki+2k$ for some $0 \leq i < k$, and for each such $i$, there is again only one possibility.
The other four cases ($B$ in $S^c$) are trivially similar; the key is still that there are at most $k$ ways to add or remove a single element to pass from $T$ to $T^c$ or vice versa.
So now we're done; choose whichever of $S$ or $S^c$ has cardinality $2^{n-1} + 1$, and it has max(max indegree, max outdegree) = $k = \sqrt{n}/\sqrt{2}$.
