For what value of $a$ does the equation $(a^2 - a - 2)x^2 + (a^2 - 4)x + a^2 - 3a + 2 = 0$ have more than two solutions?

Consider the following two statements:
$I$. Any pair of consistent linear equations in two variables must have a unique solution.
$II$. There do not exist two consecutive integers,the sum of whose squares is $365$.
Then,

The minimum value of $f(x) = x^2 + 4x + 5$ is . . . . . . ,where $x \in R$.

If the difference between the roots of the equations $x^2 - bx + c = 0$ and $x^2 - cx + b = 0$ is the same,then $b + c = \dots$

If $\alpha$ and $\beta$ $(\alpha < \beta)$ are the roots of the equation $(-2+\sqrt{3})(|\sqrt{x}-3|) + (x-6\sqrt{x}) + (9-2\sqrt{3}) = 0$, $x \ge 0$, then $\sqrt{\frac{\beta}{\alpha}} + \sqrt{\alpha\beta}$ is equal to:

The least integer $k$ which makes the roots of the equation $x^2 + 5x + k = 0$ imaginary is