3 Bulgarian Spring Tournament

a) Since ${\log }_{\frac{1}{2}}(a^4)=-2{\log }_2(a^2)$, then by putting $2{\log }_2(a^2)=b$, we get the inequality $x^2+b\cdot x+3+b<0$. For this inequality to have at least one solution, it is necessary and sufficient that $D=b^2-4b-12>0$ whose solutions are $b<-2$ or $b>6$, whence ${\log }_2(a^2)<-1$ or ${\log }_2(a^2)>3$. From the properties of the logarithmic function, we get $a^2<\frac{1}{2}$ or $a^2>8$ and $a\ne 0$. Final $$a\in (-\infty ;-2\sqrt{2})\cup (-\frac{\sqrt{2}}{2};0)\cup (0;\frac{\sqrt{2}}{2})\cup (2\sqrt{2};\infty ).$$

b) Since $a<0$, we get: $$(1)\sqrt{a^2-a+1}+a=\frac{(\sqrt{a^2-a+1}+a)(\sqrt{a^2-a+1}-a)}{\sqrt{a^2-a+1}-a}=\frac{-1+\frac{1}{a}}{-\sqrt{1-\frac{1}{a}+\frac{1}{a^2}}-1}.$$ Therefore $$\underset{a\to -\infty }{\lim }\left(\sqrt{a^2-a+1}+a\right)=\frac{1}{2}.$$