By the Cayley-Hamilton theorem, $A$ satisfies $A^2-(\text{tr}A)A+(\det A)I=O$. Comparing with $A^2-4A+I=O$ gives $\text{tr}A=4$, hence $1+\alpha =4\Rightarrow \alpha =3$.

Similarly $B^2-(\text{tr}B)B+(\det B)I=O$ matched with $B^2-5B-6I=O$ gives $\det B=-6$, so $6-3\beta =-6\Rightarrow \beta =4$.

Therefore $B-A=\left[\begin{array}{c}21\\ 3-1\end{array}\right]$, $B+A=\left[\begin{array}{c}45\\ 55\end{array}\right]$, and

$(B-A)(B+A)=\left[\begin{array}{c}1315\\ 710\end{array}\right]$

The transpose is $\left[\begin{array}{c}137\\ 1510\end{array}\right]$, which is not the matrix in $(S_1)$. So $(S_1)$ is false.

For $(S_2)$: $A+B=\left[\begin{array}{c}45\\ 55\end{array}\right]$, $\det (A+B)=20-25=-5$. For a $2\times 2$ matrix, $\det (\text{adj}M)=\det M$, so $\det (\text{adj}(A+B))=-5$. $(S_2)$ is correct.

Hence only $(S_2)$ holds.