Помогите решить алгебру

0}} \right.\\\\\left \{ {{-x<-2} \atop {-x>-3}} \right.\\\\\left \{ {{x>2} \atop {x<3}} \right." alt="1)Cos^{2} 22,5^{o}-Sin^{2}22,5^{o}=Cos(2*22,5)^{o}=Cos45^{o}=\frac{\sqrt{2} }{2}\\\\2)5^{log_{5} (3-x)} <1\\\\\left \{ {{3-x<1} \atop {3-x>0}} \right.\\\\\left \{ {{-x<-2} \atop {-x>-3}} \right.\\\\\left \{ {{x>2} \atop {x<3}} \right." align="absmiddle" class="latex-formula">

Ответ : x ∈ (2 ; 3)

$3)\int\limits^4_2 {x^{3} } \, dx=\frac{1}{4}x^{4}|_{2}^{4}=\frac{1}{4}(4^{4}-2^{4})=\frac{1}{4}(256-16)=60$

4) y = 3ˣ + 4

3ˣ > 0 при любых значениях x , значит 3ˣ + 4 > 4

Область значений : y ∈ (4 ; + ∞)