$a)$  

Zaczynamy od wyznaczenia dziedziny: 

$\{\begin{array}{l}2x+5\ge 0\\ x-1\ge 0\end{array}\Rightarrow \{\begin{array}{l}2x\ge -5\\ x\ge 1\end{array}\Rightarrow \{\begin{array}{l}x\ge -\frac{5}{2}\\ x\ge 1\end{array}\Rightarrow x\in ⟨1,+\infty )$  

$\sqrt{2x+5}+\sqrt{x-1}=8{\mid }^2$  

$2x+5+2\cdot \sqrt{2x+5}\cdot \sqrt{x-1}+x-1=64$  

$3x+4+2\sqrt{\left(2x+5\right)\cdot \left(x-1\right)}=64\mid -3x-4$  

$2\sqrt{\left(2x+5\right)\cdot \left(x-1\right)}=60-3x{\mid }^2$  

$4\left(2x+5\right)\left(x-1\right)=3600-360x+9x^2$