You have no idea how to solve this? Even if you are a beginner, if you are given a problem like this, you should know how to find eigenvalues- at least the definition of "eigenvalue".
A number, $\displaystyle \lambda$, is said to be an "eigenvalue" of linear transformation A if and only if there exist a non-zero ("non-trivial") vector, v, satisfying $\displaystyle Av= \lambda v$. Obviously, v= 0 always satisfies that- the key here is "non-zero".
That equation is the same as $\displaystyle Av- \lambda v= (A- \lambda I)v= 0$. Now, in terms of matrices, if $\displaystyle A- \lambda I$ had an inverse, we could multiply on both sides by that inverse, $\displaystyle (A- \lambda I)^{-1}(A- \lambda I)v= v= (A- \lambda I)0= 0$ showing that v= 0, the "trivial" solution is the only solution.
So $\displaystyle \lambda$ is an eigenvalue of A if and only if $\displaystyle A- \lambda I$ does not have an inverse- and that is true if and only if the determinant of $\displaystyle A- \lambda I$ (written as a matrix) is 0. That is, any eigenvalue, $\displaystyle \lambda$ must satisfy the "characteristic equation" $\displaystyle \left|A- \lambda I\right|= 0$
Here, $\displaystyle A= \begin{pmatrix}3 & 1 \\ -1 & -\mu\end{pmatrix}$ so $\displaystyle A- \lambda I= \begin{pmatrix}3 & 1 \\ -1 & -\mu\end{pmatrix}- \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}= \begin{pmatrix}3-\lambda & 1 \\ -1 & -\mu-\lambda\end{pmatrix}$
The "characteristic equation is
$\displaystyle \left|\begin{array}{cc}3-\lambda & 1 \\ -1 & \mu-\lambda\end{array}\right|= (3- \lambda)(\mu- \lambda)+ 1-= \lambda^2- (\mu+ 3)\lambda+ 1+ 3\mu= 0$.
That is a quadratic equation to be solved for $\displaystyle \lambda$. Using the quadratic formula to solve it will let you determine what values of $\displaystyle \mu$ give two real, one real, or two complex solutions.