Given X is a symmetrical random variable and $E[X^4]$ exists. Find $a$ such that $E[(X-a)^4]$ is minimised.
The fact, X is symmetrical, means the odd central moments are $0$. I expanded the expression, worked out the derivative and set it to $0$. I ended up with an answer of $a^2 = -3E[X^2]$. I'm not sure if this is correct.
Your calculation is (almost) correct.
\begin{align} E[(X-a)^4] &= a^4 + 4a^3E[X]+ 6a^2E[X^2] + 4a^3E[X] + E[X^4] \\ &= a^4 + 6a^2E[X^2] + E[X^4] \end{align}
because, as you mention yourself, "the odd central moments are 0", i.e., $$ E[X] = E[X^3] = 0 $$
and thus
$$ \frac{d}{da}E[(X-a)^4] = 4a^3 + 12aE[X^2] $$
This is $0$ only when $a=0$ (the other 2 roots of the cubic polynomial are the complex conjugates you found).
Since the derivative changes sign when $a=0$, this is indeed, an extreme value, and since the change is from negative (function is decreasing) to positive (function is increasing), this is a minimum.
Thus the answer is $a=0$.