(i) If $A$ is skew-symmetric, that is $A^T = −A$, then $\mathrm{tr}(A) = 0.$
True because a skew-symmetric matrix have $0$s along the diagonal.
(ii) If $n$ is odd and $A$ is skew-symmetric, then $\det(A) = 0$.
True for same reason as above.
(iii) If $A^3 = I_n$, then $\det(A) = 1$.
$A = I_n*A^{-2}$ hence $\det(A) = \det(I_n)*\det(A^{-2}) = \det(A^{-2})$ which is not necessarily 1.
(iv) If $A^2 = A$, then $A$ is singular.
$A^{-1} = A$ but as $A$ is singular the inverse doesn't exist so contradiction??
(v) If $n = 2$, then $A^2 − \mathrm{tr}(A)A + \det(A)I_2 = A^2 − \mathrm{tr}(A)A − \frac{1}{2}\mathrm{tr}(A^2)I_2 + \frac{1}{2}\mathrm{tr}(A)^2I_2$ =0.
Is there a quick way to answer this?
(i) Is correct
(ii) The reason given is incorrect.
Note that for any matrix, $\det(A) = \det(A^T)$, if $n$ is odd and $A^T = - A$, then $\det(A)=\det(A^T)=\det(-A)=(-1)^n \det(A) = -\det(A)$. The only number $x$ such that $x=-x$ is $x=0$.
(iii) See @tilper's comment, and @Thomas Andrews.
It is actually true if $A$ is restricted to have real entries. Because $A^3=I_n$ means $\det(A)^3=\det(I_n)=1,$ which means $\det(A)=1$, so $\det(A)=e^{2πi/3}$, or $\det(A)=e^{4πi/3}$. The latter two aren't possible if $A$ has real entries.
(iv) This is false, as @user40085 points out. $I_n$ satisfies the hypotheses but has non-zero determinant.
(v) If $n=2$, then $A^2−tr(A)A+\det(A)I_2 =0$.
This is true, see the Cayley-Hamilton theorem and Vieta's formulas.