1. 
2. 
0 " alt=" \frac{D}{4}=(a-1)^2 -a(a-4)>0 " align="absmiddle" class="latex-formula">
0\\ 2a+1>0\\ a>-0.5 " alt=" a^2-2a+1-a^2+4a>0\\ 2a+1>0\\ a>-0.5 " align="absmiddle" class="latex-formula">
Корни уравнения: 
Согласно условию: 
3 " alt=" \displaystyle \bigg|x_2-x_1\bigg|=\bigg|-\frac{2\sqrt{2a+1}}{a}\bigg|>3 " align="absmiddle" class="latex-formula">
3|a|\\ \\ \left \{ {{2a+1\geq0} \atop {4(2a+1)>9a^2}} \right. ~~~\Rightarrow~~~\left \{ {{a\geq-0.5} \atop {9a^2-8a-4<0}} \right. ~~~\Rightarrow~~~\left \{ {{a\geq-0.5} \atop {a\in(\frac{4-2\sqrt{13}}{9}};\frac{4+2\sqrt{13}}{9})} \right. " alt=" \displaystyle 2\sqrt{2a+1}>3|a|\\ \\ \left \{ {{2a+1\geq0} \atop {4(2a+1)>9a^2}} \right. ~~~\Rightarrow~~~\left \{ {{a\geq-0.5} \atop {9a^2-8a-4<0}} \right. ~~~\Rightarrow~~~\left \{ {{a\geq-0.5} \atop {a\in(\frac{4-2\sqrt{13}}{9}};\frac{4+2\sqrt{13}}{9})} \right. " align="absmiddle" class="latex-formula">
Общее пересечение решения неравенств: 