, 2022

For simplicity we write $A$ instead of Ivan and $B$ instead of Peter. The crucial idea is in the following

Lemma. In the graph $G$ all vertices are initially colored white. Let $A$ colors some of the vertices in black and $B$ asks $A$ about the parity of the number of blacks in a set $S$ from vertices, and with the answer can unambiguously determine the parity of the number of differently colored pairs adjacent/vertices in $G$. Then $S$ must be the set of all vertices of odd degree in $G$.

Proof. We will work by modulus $2$. We write the number $1$ in each black vertex, the number $0$ in each white vertex and the sum of the numbers in its vertices on each edge. Each black vertex of odd degree contributes $1$ to the sum of the numbers on the edges, each black vertex of even degree contributes $0$, and each white vertex also contributes $0$. Therefore, the sum of the numbers on the edges in $G$ has the same parity as that of the number of black vertices of odd degree in $G$. Thus, when asked about the set $N$ of the vertices of odd degree, we are ready. Conversely, for a question with a set other than $N$, we cannot unambiguously restore the parity of the number of black vertices in $N$ and the lemma is proved.

It remains to count the number of cells with an odd number of adjacent cells in $T$. Each cell that has no common point with the perimeter of $T$ has $12$ adjacent cells. The three cells on the perimeter, each of which is adjacent on the side of a corner cell in $T$, have $6$ adjacent cells. Each cell with a side on the perimeter has $7$ adjacent cells, and each cell with a vertex (but not a side) on the perimeter (except the three mentioned above) has $9$ adjacent cells. Because each side of $T$ has at least one common point with $2 \times 2022 - 1$ cells, and each two sides have two common cells (as only one is of odd degree), the required number is $3 \times (2 \times 2022 - 5) + 3 = 12120$.