Please forgive me if I use a decimal point instead of a decimal comma; I have a hard time using the comma...
First, I don't see how you got from the first to the second line.
Second, to go from the second to the third line, you seem to have divided both sides by $0.5$ to get $6\leq x^2$, and then you tried taking the square root. Unfortunately, you seem to have forgotten that $\sqrt{x^2}=|x|$, not $x$. so you should really have gotten $\sqrt{6}\leq |x|$, which would have let to $x\geq \sqrt{6}$ or $x\leq -\sqrt{6}$.
Okay, let's take it from the top. First, let us "bring down" the exponent by taking logarithms. Since the logarithm is a stricitly increasing function, $0\lt a\leq b$ holds if and only if $\ln(a)\leq \ln(b)$ holds. So $\begin{align*} (0.25)^{3-0.5x^2} &\leq 8\\ (3-0.5x^2) \ln(0.25) &\leq \ln(8)\\ -0.5x^2\ln\left(\frac{1}{4}\right) &\leq \ln(8) -3\ln\left(\frac{1}{4}\right)\\ -0.5x^2(-\ln(4)) &\leq \ln(8) + 3\ln(4)\\ x^2\ln(2) &\leq 3\ln(2)+6\ln(2)\\ x^2\ln(2) &\leq 9\ln(2)\\ x^2 &\leq 9. \end{align*}$ The last inequality because $\ln(2)\gt 0$, so dividing through by $\ln(2)$ does not affect the inequality sign. Now we can take square roots on both sides and we get $|x|=\sqrt{x^2} \leq 3$ and this is equivalent to $-3\leq x\leq 3$; i.e., to the solution set being $[-3,3]$.