Find all 2 by 2 complex matrix with the following condition 
Find all $A \in \mathrm{Mat}_{2 \times 2} ( \mathbb{C})$ such that
$A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
$A$ with $A^2 = -I$. Deduce that for every even $n$ there are
infinitely many $n \times n$ real matrices with $A^2 = -I$.

Try:
We have
$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies A^2 = \begin{bmatrix}a^2+bc & b(a+d) \\ c(a+d) & cb+d^2 \end{bmatrix} $$
So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$
If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.
if $a^2=-1$, then $a= \pm i$ and similarly $b= \pm i$. As long as $a+d \neq 0$, we see that $c=0$ otherwise $c \in \mathbb{C}$ is arbitrary and so we have
$$ \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}, \begin{bmatrix} -i & 0 \\ 0 & -i \end{bmatrix},\begin{bmatrix} -i & 0 \\ c & i \end{bmatrix},\begin{bmatrix} i & 0 \\ c & -i \end{bmatrix} $$
now if $b \neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$
Therefore, we have
$$ \begin{bmatrix} a & b \\ -(1+a^2)/b & -a \end{bmatrix} $$
are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b  \neq 0$ and so we get matrices like the one above only.
Now, is this correct so far? I'm stuck for the $n\times n$ case. Any help?

Looking very good so far. However, for the $a=\pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.
For even $n$, take the real $2\times2$ matrix you've found, make $\frac n2$ copies of it all along the diagonal of an $n\times n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $n\times n$ matrices.
Bonus question: What happens for odd $n$ with real matrices?
