Solution of Logarithmic Inequalities: $\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1$ 
If $\log_{0.5}(\log_5 (x^2-4)) >\log_{0.5}1$ then x lies in the interval:
  (a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3)$
  (a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3\sqrt{5})$
  (c) $(\sqrt{5}, 3\sqrt{5})$
  (d) $\phi $    

I have solved quite a few logarithmic inequalities and I am familiar with the procedure. I was just not able to get the answer that matches with the given options in the question.   
My Working:
$\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1 $
$\log_{5}(x^2-4)<1$
$\log_{5}(x^2-4)<\log_5(5)$
$x^2-4<5$
$x^2<9$    
$\therefore -3<x<3$    
Now the second equation comes from the fact that
$x^2-4>0$
$\therefore x<-2, x>2$   
The intervals formed are $(-3,-2)\cup(2,3)$

First of all, $\log_{0.5}(\log_5 (x^2-4))$ will remain defined iff 
$\log_5 (x^2-4)>0\iff x^2-4>5^0=1\iff x^2>5\iff$ either $x>\sqrt5$ or $x<-\sqrt5\  \ \ \  (1)$
$$\log_{0.5}(\log_5(x^2-4))>\log_{0.5}1\iff\dfrac{\log_e(\log(x^2-5))}{\log_e(0.5)}>\dfrac{\log1}{\log_e(0.5)}=0$$
Now as $0.5=\dfrac12=2^{-1},\log_e(0.5)=\log_e(2^{-1})=-\log_e2$
$\implies0>\dfrac{\log_e(\log_5(x^2-4))}{\log_e(0.5)}=-\dfrac{\log_e(\log_5(x^2-4))}{\log_e2}$
$\implies\log_e(\log_5(x^2-4))<0\iff\log_5(x^2-4)<e^0=1$ 
$\iff x^2-4<5^1=5\iff x^2<5+4\iff-3<x<3 \  \ \ \  (2)$
What is the intersection of $(1),(2)?$
