1. **State the problem:** We need to find the $(1,2)$-entry of matrix $A$ given that $$(2A^{-1}+I)^{-1} = \begin{bmatrix}3 & 4 \\ 5 & 7\end{bmatrix}.$$ 2. **Rewrite the equation:** Let $M = 2A^{-1} + I$. Then we have $$M^{-1} = \begin{bmatrix}3 & 4 \\ 5 & 7\end{bmatrix}.$$ Taking the inverse on both sides gives $$M = \left(\begin{bmatrix}3 & 4 \\ 5 & 7\end{bmatrix}\right)^{-1}.$$ 3. **Find $M$ by inverting the given matrix:** For a $2\times 2$ matrix $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$, the inverse is $$\frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}.$$ Calculate the determinant: $$ad - bc = 3 \times 7 - 4 \times 5 = 21 - 20 = 1.$$ So, $$M = \begin{bmatrix}7 & -4 \\ -5 & 3\end{bmatrix}.$$ 4. **Express $M$ in terms of $A^{-1}$:** Recall $$M = 2A^{-1} + I,$$ so $$2A^{-1} = M - I = \begin{bmatrix}7 & -4 \\ -5 & 3\end{bmatrix} - \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}6 & -4 \\ -5 & 2\end{bmatrix}.$$ 5. **Solve for $A^{-1}$:** $$A^{-1} = \frac{1}{2} \begin{bmatrix}6 & -4 \\ -5 & 2\end{bmatrix} = \begin{bmatrix}3 & -2 \\ -\frac{5}{2} & 1\end{bmatrix}.$$ 6. **Find $A$ by inverting $A^{-1}$:** The determinant of $A^{-1}$ is $$3 \times 1 - (-2) \times \left(-\frac{5}{2}\right) = 3 - 5 = -2.$$ The inverse is $$A = \frac{1}{-2} \begin{bmatrix}1 & 2 \\ \frac{5}{2} & 3\end{bmatrix} = \begin{bmatrix}-\frac{1}{2} & -1 \\ -\frac{5}{4} & -\frac{3}{2}\end{bmatrix}.$$ 7. **Identify the $(1,2)$-entry of $A$:** The entry in the first row and second column is $$-1.$$ **Final answer:** The $(1,2)$-entry of $A$ is **$-1$**.