Problem 1381

a) Since $\log_{\frac{1}{2}}(a^4) = -2 \cdot \log_2(a^2)$, 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 $a \neq 0$. Final
$$a \in (-\infty; -2\sqrt{2}) \cup \left(-\frac{\sqrt{2}}{2}; 0\right) \cup (0; \frac{\sqrt{2}}{2}) \cup (2\sqrt{2}; \infty).$$

b)
$$(1) \quad \sqrt{a^2 - a + 1} + a = \frac{(\sqrt{a^2 - a + 1} + a)(\sqrt{a^2 - a + 1} - a)}{\sqrt{a^2 - a + 1} - a} = \frac{-1 + \frac{1}{a}}{-\sqrt{1 - \frac{1}{a} + \frac{1}{a^2}} - 1}.$$
Therefore
$$\lim_{a \to -\infty} (\sqrt{a^2 - a + 1} + a) = \frac{1}{2}.$$

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:
$$(1) \qquad \sqrt{a^2 - a + 1} + a = \frac{(\sqrt{a^2 - a + 1} + a)(\sqrt{a^2 - a + 1} - a)}{\sqrt{a^2 - a + 1} - a} = \frac{-1 + \frac{1}{a}}{-\sqrt{1 - \frac{1}{a} + \frac{1}{a^2}} - 1}.$$
Therefore
$$\lim_{a \to -\infty} \left( \sqrt{a^2 - a + 1} + a \right) = \frac{1}{2}.$$