This question can be easily done by using the concept of characteristic equation of a matrix.
Consider the equation $\text{det (A - kI)}=0$, where k is a scalar.
Now, when we solve the equation $\left|\text{A}-\text{k}\text{I}\right|=0$, then we get a polynomial in k.
Here, if we put k = A(matrix), then the equation obtained is called as the characteristic equation of square matrix A.
The characteristic polynomial of A is $\left|\text{A}-\text{k}\text{I}\right|=0$.
$\therefore \left|\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right]-\text{k}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$
$\Rightarrow \left|\left[\begin{array}{cc}3-\text{k}& 2\\ 7& 5-\text{k}\end{array}\right]\right|=0$
$\Rightarrow \left|\begin{array}{cc}3-\text{k}& 2\\ 7& 5-\text{k}\end{array}\right|=0$
$\Rightarrow \left(3-\text{k}\right)\left(5-\text{k}\right)-14=0$
$\Rightarrow {\text{k}}^2-8\text{k}+1=0$
$\therefore {\text{A}}^2-8\text{A}+\text{I}=0$
On comparing the above equation with the given equation ${\text{A}}^2+\lambda \text{A}+\mu \text{I}=\text{O}$, we get
$\lambda =-8,\mu =1\Rightarrow \frac{\mu -\lambda }{3}=3$