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The answer to Question 1 is $52!/48! - 4 \binom{13}{2} \binom{50}{2}= 6115200$.

That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, note that a $4$-tuple $(a,b,c,d)$ cannot occur if and only if the fifth card $e$ that $(a,b,c,d)$ signifies is among $\{b,c,d\}$. Note that there are $4 \binom{13}{2}$ choices for $(a,e)$. Finally, for each of the $\binom{50}{2}$ $2$-subsets $\{x,y\}$ of $[52] \setminus \{a,e\}$, there is a unique ordering $(b,c,d)$ of $\{ x,y,e \}$ such that $(a,b,c,d)$ signifies $e$.

As Ilya Bogdanov points out in the comments, a $5$-set has a unique $4$-tuple if and only if it contains exactly two cards of the same suit. The number of such $5$-sets is $4 \binom{13}{2}13^3$. Thus, the answer to Question 2 is $\binom{52}{5}-4 \binom{13}{2}13^3= 1913496$.

The answer to Question 1 is $52!/48! - 4 \binom{13}{2} \binom{50}{2}= 6115200$.

That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, note that a $4$-tuple $(a,b,c,d)$ cannot occur if and only if the fifth card $e$ that $(a,b,c,d)$ signifies is among $\{b,c,d\}$. Note that there are $4 \binom{13}{2}$ choices for $(a,e)$. Finally, for each of the $\binom{50}{2}$ $2$-subsets $\{x,y\}$ of $[52] \setminus \{a,e\}$, there is a unique ordering $(b,c,d)$ of $\{ x,y,e \}$ such that $(a,b,c,d)$ signifies $e$.

As Ilya Bogdanov points out in the comments, a $5$-set has a unique $4$-tuple if and only if it contains exactly one pair of cards of the same suit. The number of such $5$-sets is $4 \binom{13}{2}13^3$. Thus, the answer to Question 2 is $\binom{52}{5}-4 \binom{13}{2}13^3= 1913496$.

I think the answer to Question 1 is $52!/48! - 4 \binom{13}{2} \binom{50}{2}= 6115200$.

That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, note that a $4$-tuple $(a,b,c,d)$ cannot occur if and only if the fifth card $e$ that the $4$-tuple signifies is among $\{b,c,d\}$. Note that there are $4 \binom{13}{2}$ choices for $(a,e)$. Finally, for every $2$-subset $\{x,y\}$, of $[52] \setminus \{a,e\}$, there is a unique ordering $(b,c,d)$ of $\{ x,y,e \}$ such that $(a,b,c,d)$ signifies $e$.

The answer to Question 1 is $52!/48! - 4 \binom{13}{2} \binom{50}{2}= 6115200$.

That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, note that a $4$-tuple $(a,b,c,d)$ cannot occur if and only if the fifth card $e$ that $(a,b,c,d)$ signifies is among $\{b,c,d\}$. Note that there are $4 \binom{13}{2}$ choices for $(a,e)$. Finally, for each of the $\binom{50}{2}$ $2$-subsets $\{x,y\}$ of $[52] \setminus \{a,e\}$, there is a unique ordering $(b,c,d)$ of $\{ x,y,e \}$ such that $(a,b,c,d)$ signifies $e$.

As Ilya Bogdanov points out in the comments, a $5$-set has a unique $4$-tuple if and only if it contains exactly two cards of the same suit. The number of such $5$-sets is $4 \binom{13}{2}13^3$. Thus, the answer to Question 2 is $\binom{52}{5}-4 \binom{13}{2}13^3= 1913496$.

I think the answer to Question 1 is $52 \cdot 51 \cdot 50 \cdot 49 - 4 \binom{13}{2} \binom{50}{2}= 6115200$.

That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, note that a $4$-tuple $(a,b,c,d)$ cannot occur if and only if the fifth card $e$ that the $4$-tuple signifies is among $\{b,c,d\}$. Note that there are $4 \binom{13}{2}$ choices for $(a,e)$. Finally, for every $2$-subset $\{x,y\}$, of $[52] \setminus \{a,e\}$, there is a unique ordering $(b,c,d)$ of $\{ x,y,e \}$ such that $(a,b,c,d)$ signifies $e$.

I think the answer to Question 1 is $52!/48! - 4 \binom{13}{2} \binom{50}{2}= 6115200$.

That is, I claim that the number of $4$-tuples that cannot occur is $4 \binom{13}{2} \binom{50}{2}$. To see this, note that a $4$-tuple $(a,b,c,d)$ cannot occur if and only if the fifth card $e$ that the $4$-tuple signifies is among $\{b,c,d\}$. Note that there are $4 \binom{13}{2}$ choices for $(a,e)$. Finally, for every $2$-subset $\{x,y\}$, of $[52] \setminus \{a,e\}$, there is a unique ordering $(b,c,d)$ of $\{ x,y,e \}$ such that $(a,b,c,d)$ signifies $e$.