$f(x+1/2)+f(x-1/2)= f(x)$ Then the period of $f(x)$ is? 
$f(x+1/2)+f(x-1/2)= f(x)$. Then the period of $f(x)$ is:
a)$1$
b)$2$
c)$3$
d)$4$?

Attempt: 
I substituted $x= x \pm1/2$ but the equations I got didn't help at all. 
How do I go about solving such a question? I am just looking for a hint and not the entire solution.

We may conclude that $f(x)=f(x+3)$ holds for all $x$.
Following @Mike Earnest's suggestion,
\begin{align}
f(x)&=f(x+1/2)+f(x-1/2),\\
f(x-1/2)&=f(x)+f(x-1).
\end{align}
Add up these two equations, and you will have
$$
f(x+1/2)+f(x-1)=0,
$$
or, due to the arbitrariness of $x$,
$$
f(x+3/2)+f(x)=0.
$$
Now let $x\to x+3/2$, and
$$
f(x+3)+f(x+3/2)=0.
$$
The difference of the last two equations gives
$$
f(x)=f(x+3).
$$
Of course, as @Przemysław Scherwentke has mentioned, this might not mean that $3$ is the period of $f$, as we do not know if this $3$ is the smallest non-negative value $T$ that ensures $f(x)=f(x+T)$. All in all, this is what we can get from the given relation.
