If a real $2\times2$ matrix satisfies $A^4=I$, does it follow that $A^2=\pm I$? I would like to know if a  $2 \times 2$ matrix $A$ satisfies $A^4=I$, also satisfies $A^2=I$ or $A^2=-I$ if all elements of $A$ are real. 
Also, I would really appreciate your help on further generalizations. 
i.e) Properties of $A$ such that $A^n=I$.
 A: Let $A^4=I$, and let $B=A^2$. Then, $B^2=I$. We can use determinants to learn a little more about B: $$\det(B^2)=\det(B)^2=\det(I)=1,$$ so $\det(B)=\pm 1$. Furthermore, since $\det(B)=\det(A^2)=\det(A)^2$, $\det(B)$ is the square of a real number and cannot be negative, so $\det(B)=1$. Thus, $B$ is invertible, so $B=B^{-1}$. It is known that every self-inverse matrix is diagonalizable, so  there exists $P$ such that $B=PDP^{-1}$ for some diagonal matrix $D$. This means that $$B^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}=I$$ and $$PD^2=P$$ so $$D^2=I.$$ Thus, if $$D=\left(\begin{array}{cc} d_1 & 0\\ 0 & d_2\end{array}\right),$$ then $d_1^2=d_2^2=1$. Now consider $\det(PDP^{-1})=\det(D)=\det(B)=1$, so $d_1$ and $d_2$ must have the same sign, so $D=\pm I$. This leaves us with $$A^2=B=PDP^{-1}=\pm PIP^{-1}=\pm I.$$
In general, if $A^n=I$, we must have that $\det(A)^n=1$. If, as you say, the entries in $A$ are real, so is the determinant of $A$, so $\det(A)=\pm 1$. If $n$ is odd, note that $\det(A)=1$. If $n$ is even, we get that $A^{n/2}=\pm I$ by the same proof given above.
A: For your first question, as you haven't learnt minimal polynomial, here is a slightly lengthier answer. Since $A^4=I$, the determinant of $A$ --- being a real number --- must be either $-1$ or $1$. Therefore, if we put $B=A^2$, then $B^2=I$ and $\det(B)=\det(A)^2=1$. Now, one can easily verify that
$$
B^2=\operatorname{trace}(B)B-\det(B)I\tag{1}
$$
by directly comparing both sides of $(1)$ entrywise. (This is actually a special case of Cayley-Hamilton theorem, but I guess you haven't learnt it yet.) So, with our $B$, we get $I=\operatorname{trace}(B)B-I$ or $\operatorname{trace}(B)B=2I$. It follows that $B=kI$ for some real number $k$. However, since $B^2=I$, $k$ must be equal to $\pm1$. Therefore $A^2=B=\pm I$.
For your second question, the answer depends on what generalization you have in mind. If you are asking whether $A^2$ must be $\pm I$ when $A^n=I$, the answer is no. Here is a counter example:
$$
A=\pmatrix{0&-1\\ 1&-1},\ A^3=I,\ A^2=\pmatrix{-1&1\\ -1&0}\ne\pm I.
$$
A: If $A^{2}\ne I$, then there is a non-zero vector $X$ such that $Y=(A^{2}-I)X \ne 0$. Then $(A^{2}+I)Y=0$. If $A^{2}\ne -I$, then $Y=(A^{2}+I)X \ne 0$ for some $X$, which gives $(A^{2}-I)Y=0$. Combining these, you get $Y_1,Y_2$ such that $A^{2}Y_1=Y_1$ and $A^{2}Y_2=-Y_2$. Using this as a basis,
$$
   A^{2}=P^{-1}\left[\begin{array}{cc}1 & 0\\0 & -1\end{array}\right]P.
$$
That's not possible because $\det(A^{2})=\det(A)^{2} \ne -1$. So, either $A^{2}=I$ or $A^{2}=-I$.
