A problem with eigenvalues 
Let A be real square matrix.
1) If $|A^3 - I|=0$, then 1 is eigenvalue of A. (false)   
2) If
  $|A|=0$, then  1 is eigenvalue of $(I+A)^2$ (true)

My solution uses the fact that if matrix M has eigenvalue $\lambda$, then $M^n$ has eigenvalue $\lambda^n$.
1) 1 is eigenvalue of $A^3$. Then A should have 1 as its eigenvalue. The problem is that (1) is supposed to be false.
2) On the one hand, 0 is eigenvalue of A. on the other hand, (I+A) has 1 as eigenvalue. My problem is that I don't see that 0 for A implies 1 for (I+A)
 A: *

*If $A^3-I=0$ then the polynomial $X^3-1=(X-1)(X^2+X+1)$ is divisible by the minimal polynomial $\mu_A$ of $A$. There are several possibilities for the minimal polynomial of $A$, one of which is $\mu_A=X^2+X+1$, and an example of such a matrix is
$$A=\begin{pmatrix}-1/2&\sqrt3/2\\-\sqrt3/2&-1/2\end{pmatrix},$$
which satisfies $A^3-I=0$, yet $1$ is not an eigenvalue of $A$.
To see why $A^3-I=0$, is quite easy: the characteristic polynomial of $A$ is $X^2+X+1$ (since $\text{tr}(A)=-1$ and $\lvert A\rvert=1$) hence, by Hamilton–Cayley theorem, $A^2+A+I=0$. Now $A^3-I=(A^2+A+I)(A-I)=0$.
To see that $1$ is not an eigenvalue of $A$, compute the determinant of $A-I$:
$$\lvert A-I\rvert=\left\lvert\begin{matrix}-3/2&\sqrt3/2\\-\sqrt3/2&-3/2\end{matrix}\right\rvert=3\neq0.$$

*There exists a non-zero vector $X$ in the kernel of $A$: $AX=0$. Then:
$$(I+A)^2X=\bigl(I+2A+A^2\bigr)X=X,$$
and this shows that $X$ is an eigenvector of $(I+A)^2$ associated with the eigenvalue $1$.



For $A$ in Part 1, I had in mind that a complex root of $X^2+X+1=j=\mathrm{e}^{2i\pi/3}=-\dfrac12+i\dfrac{\sqrt3}2$, a matrix representation of which is the matrix $A$ I gave. For this, it's quite handy to know that the following:
$$\mathbb{C}\longrightarrow M_2(\mathbb{R}):z=x+iy\longmapsto\begin{pmatrix}x&-y\\y&x\end{pmatrix}$$
is a ring monomorphism; the matrix on the right corresponds to the matrix of the endomorphism of the multiplication by $z$ in the complex plane identified with $\mathbb{R}^2$.

It seems from the comment that you're not familiar with the concept of minimal polynomial. So let's try a different approach. Since $\bigl\lvert A^3-I\bigr\rvert=0$, there exists a non-zero vector $X$ such that
$$\bigl(A^3-I\bigr)X=0,$$
hence (since $A^3-I=(A-I)\bigl(A^2+A+I)$):
$$(A-I)\bigl(A^2+A+I\bigr)X=0.$$
So you see that it's sufficient to have $\bigl(A^2+A+I\bigr)X=0$. We can then conclude that assertion 1 is false by exhibiting a matrix $A$ such that $A^2+A+I$ has a zero determinant.
