$ P (| X |> 1) = P (| X | <1) $

We can immediately dismiss the possibility that $a = 1$, because if that were true, then $X in [-1,1]$, from which it would follow that $Pr[|X| > 1] = 0$.

Let's translate this into something a bit more intuitive. $X$ is a number chosen at random between $-a$ and $a$, inclusive, and all values in this range are equally likely to be chosen.

If $a$ is less than $1$, then its impossible to get a value of $X$ whose absolute value will be greater than $1$, because the resulting interval is too small.

If $a$ is very large, say $1000$, then the outcome $|X| < 1$ is extremely unlikely because the range of possible values is $[-1000, 1000]$ of which $[-1,1]$ is only a tiny part.

Another way to understand the question is to think of a number line. Since we know that $a > 1$, we have the points $-a, -1, 0, 1, a$ in order from left to right. From $-a$ to $-1$, color the interval red. From $-1$ to $1$, color the interval blue. And from $1$ to $a$, color the interval red. The condition $Pr[|X| < 1] = Pr[|X| > 1]$ means that the total length of the red intervals must equal the length of the blue interval. What would $a$ need to be to make this true?