I tried to do a lot of stuff to get my answer but always fail. $$5 = -\frac{b}{2a} \Rightarrow b=-10a$$ $$\frac{1}{2}= -\frac{-b^2+4ac}{4a} \Rightarrow c=\frac{-49}{8}$$ $$\tan 45=\frac{\frac{1}{2}}{x}\Rightarrow x=\frac{1}{2} $$
Therefore I can say that $x_1=5-\frac{1}{2}=\frac{9}{2}$ and $x_x=5+\frac{1}{2}=\frac{11}{2}$
After that I really don't know what my next steps should be. I'm not even sure if these are correct anyway...
In this answer, I will focus on the cases in which the axis of the parabola is horizontal or vertical. Firstly, I consider the case where the parabola has a vertical axis, i.e. parallel to the $y $-axis. Define the equation of the parabola as $y=ax^2+bx+c $. Its derivative $2ax+b \ $ must be zero for $x=5 \ $, so we have $a=-b/10 \ $. Also, since the parabola crosses the vertex $(5,1/2) \ $, we have $1/2=-b/10 \cdot 5^2 + 5b + c \ $, which leads to $c=\frac {1}{2}- \frac {5}{2} b \ $. So the equation of the parabola can be rewritten as $$y=- \frac {b}{10} x^2 +bx + \frac {1}{2} - \frac {5}{2} b \ $$
Setting $y=0$ we get that the points in which the parabola crosses the $x $-axis are given by the two solutions of the equation $$ -\frac {b}{10} x^2 -\frac {3}{2} bx +1/2=0 \ $$ which are given by $x=5 \pm \sqrt {\frac {5}{b}} \ $. In these points, the derivative $ 2ax+b =-\frac {1}{5} bx+b \ $ must be equal to $\pm 1$ (because the parabola crosses the $x $-axis at $45^o$), so we get the two equations
$$-\frac {b}{5} (5 - \sqrt {\frac {5}{b}} ) +b=1$$ $$- \frac {b}{5} (5 + \sqrt {\frac {5}{b}} ) +b=-1$$
whose solution is $b=5$ . Then $a=-1/2$ , $c=-12$ , and the equation of the parabola is
$$y=-\frac {1}{2}x^2+5x-12 \ $$
Among the five points provided in the possible answers, only the point $(-2,-24) \ $ is crossed by this parabola.
We can also repeat the same steps for the case where the axis of the parabola is horizontal. The same procedure leads to the following two parabolas:
$$ x=y^2-y+5+1/4$$ $$ x=-y^2+y+5-1/4$$
However, in this case none of the five points provided in the possible answers is crossed by the parabolas. Lastly, we have to take into account that other potential solutions might result by considering oblique parabolas. In this case, it is likely that we could obtain multiple solutions, i.e., a set of parabolas that makes different answers (among the five provided) potentially valid.