(a) $\det(A)=1$
(b) $\det(A)=-1$
(c) $\text{Trace}(A)=0$
(d) $\text{Trace}(A)=1$
Explanation: (a) and (c) are correct.
Let $\lambda$ and $\mu$ are e-values of $A$, $\implies \frac{1}{\lambda}$ and $\frac{1}{\mu}$ are e-values of $A^{-1}$.
Given that det$(A)\neq 0$, i.e. $\lambda \neq 0, \mu \neq 0$.
Let $Ch(A) = x^2 - (\lambda+\mu)x + \lambda\mu$, and $Ch(A^{-1}) = x^2 - (\frac{1}{\lambda}+\frac{1}{\mu})x + \frac{1}{\lambda\mu}$.
(Note that $Ch(A) = x^2 - (\text{trace}(A))x + \text{det}(A)$)
Now, after some simplification we have, $Ch(A) - Ch(A^{-1}) = -(\lambda+\mu)x+\left(\frac{1}{\lambda}+\frac{1}{\mu}\right)x + \left(\lambda\mu - \frac{1}{\lambda\mu}\right).$
This will be constant if coefficient of $x$ is zero.
i.e. $-(\lambda+\mu)+\frac{1}{\lambda}+\frac{1}{\mu} = 0$.
$\implies (\lambda+\mu)\left(\frac{1-\lambda\mu}{\lambda\mu}\right) = 0$.
$\implies \lambda+\mu = 0$, or $1-\lambda\mu = 0$.
i.e. $\text{Trace}(A) = 0$, or $\det(A)=1$.
$\implies$ (a) and (c) are correct.