Помогите,пожалуйста,номер 2 "а" и 3 "б"

2а)Замена (1/3)^x=t
t^2+8t-9=0
t1=-9 t2=1
(1/3)^x=-9 (1/3)^x=1
x=-2 x=0
3б) Замена log0.5X=t
t^2-3t-4≤0
t1=-1 t2=4
t∈[-1;4]
$\left \{ {t1 \geq -1} \atop{t2 \leq 4}} \right. \left \{ {{log_{0,5}}x \geq -1} \atop {log_{0.5}x \leq 4}} \right. \left \{ {{log_{0.5}x \geq log_{0.5}2} \atop {log_{0,5}x \leq log_{0,5} \frac{1}{16} }} \right. \left \{ {{x \leq 2} \atop {x \geq \frac{1}{16} }} \right.$
x∈[0.0625;2]