Binomial expansion.

We know that $\displaystyle \left(1-\frac {x^2}4\right)^8 = \sum_{n=0}^8 (-1)^n \binom 8n \left(\frac {x^2}4\right)^n$ . Therefore, the coefficient of $x^6$ is when $n=3$ $\implies a_3 = (-1)^3 \dbinom 83 \left(\dfrac 14\right)^3 = - \dfrac {8\cdot 7 \cdot 6}{3\cdot 2}\cdot \dfrac 1{4^3} = - \dfrac 78 = \boxed{- 0.875}$ .