I use some of the same properties, particularly the "optical" property, as the other responders here, but in a somewhat different way, so some of the same calculations will appear, but in a different guise. (This seems closest to hypergeometric's approach.)
There is a "similarity" property we can apply. For the "upward-opening" parabola with vertex at the origin, $ \ y \ = \ \frac{1}{4p} \ x^2 \ $ ($ \ p \ $ being the "focal distance", the distance from the vertex to the focus or to the directrix), so points on the curve can be characterised as $ \ ( \ \pm \ 2 \ \sqrt{k} \ p \ , \ kp \ ) \ $ . The slope of the tangent line to a point of the curve is $ \ y' \ = \ \frac{1}{2p} \ x \ $ , hence the slope at this point is $ \ y' \ = \ \sqrt{k} \ $ . We cannot use this immediately, however, as we can establish fairly quickly that the symmetry/focal axis of the parabola is not parallel to either coordinate axis. So we are faced with a bit of additional work.
We will make use of the "optical" property that the angle which a line parallel to the symmetry/focal axis makes to the tangent line to a point on the parabola is congruent to the angle that a line from the focus to that point makes to the same tangent line. Since we know nothing about the orientation of the parabola, we might use a scalar product of vectors to determine something about this angle. The vector from the focus $ \ ( -1, \ -1 ) \ $ to the given tangent point $ \ (7, \ 13) \ $ is $ \ \langle \ 8, \ 14 \ \rangle \ $ and the slope of the tangent line is $ \ 3 \ $ , which we may represent by a vector $ \ \langle \ 1, \ 3 \ \rangle \ $ . So we can compute
$$ \cos \theta \ \ = \ \ \frac{\langle \ 8, \ 14 \ \rangle \ \cdot \ \langle \ 1, \ 3 \ \rangle }{ ( \sqrt{8^2 \ + \ 14^2} ) \ (\sqrt{1^2 \ + \ 3^2} ) } \ \ = \ \ \frac{50}{ \sqrt{260} \ \cdot \ \sqrt{10} } \ \ = \ \ \frac{5}{\sqrt{26}} \ . $$
We will similarly represent the slope of the symmetry axis by a vector $ \ \langle \ 1, \ M \ \rangle \ $ . We want the same acute angle between this vector and that for the tangent line, giving us
$$ \cos \theta \ \ = \ \ \frac{\langle \ 1, \ 3 \ \rangle \ \cdot \ \langle \ 1, \ M \ \rangle }{ ( \sqrt{10} ) \ (\sqrt{1^2 \ + \ M^2} ) } \ \ = \ \ \frac{1 \ + \ 3M}{ ( \sqrt{10} ) \ (\sqrt{1^2 \ + \ M^2} ) } \ \ = \frac{5}{\sqrt{26}} \ . $$
This can be re-arranged into the quadratic equation $$ \ 16 \ M^2 \ - \ 156 \ M \ + \ 224 \ = \ 4 \ M^2 \ - \ 39 \ M \ + \ 56 \ = \ 0 \ \ , $$
with the solutions $ \ M \ = \ \frac{39 \ \pm \ \sqrt{625}}{8} \ \ = \ \ 8 \ \ , \ \ \frac{7}{4} \ $ . We want the steeper of these slopes, giving us the slope of the symmetry axis as $ \ M \ = \ 8 \ $ . (From this, we can develop some of mathlove's results, though we won't have need of those here.)
We see from this that the parabola is rotated "off the vertical" (by slightly over 7º clockwise, as it turns out), so we need to find the slope of the given tangent line relative to the symmetry axis in order to apply the aforementioned similarity property. The angle which the tangent line makes to the "horizontal" axis is given by $ \ \tan \phi \ = \ 3 \ $ (from its stated slope), so in a coordinate system for which the symmetry axis is "vertical" , we find the "transformed" slope from the "angle-addition formula" for tangent as
$$ \ \tan \phi \ ' \ \ = \ \ \frac{3 \ + \ \frac{1}{8}}{1 \ - \ 3 \ \cdot \ \frac{1}{8}} \ \ = \ \frac{25}{5} \ \ = \ \ 5 \ \ , $$
the $ \ \frac{1}{8} \ $ coming from the tangent of the clockwise angle that we must rotate the coordinate axes, which is the cotangent of the angle that the symmetry axis makes to the horizontal axis. The similarity property then tells us that $ \ y' \ = \ \sqrt{k} \ = \ 5 \ \Rightarrow \ k \ = \ 25 \ $ .
We still require the perpendicular distance of the tangent point from the symmetry axis. The line normal to that axis through $ \ (7, \ 13) \ $ is $ \ y \ - \ 13 \ = \ -\frac{1}{8} \ ( x \ - \ 7 ) \ \Rightarrow \ y \ = \ \frac{111}{8} \ - \ \frac{1}{8} \ x \ $ . The equation of the line containing the symmetry/focal axis is $ \ y \ + \ 1 \ = \ 8 \ ( x \ + \ 1 ) $ $ \Rightarrow \ \ y \ = \ 8 \ x \ + \ 7 \ $ , and these two lines intersect at $ \ \left( \frac{11}{13} \ , \ \frac{179}{13} \right) \ $ (as mathlove also finds).
At last, we apply the similarity property: the perpendicular distance from the symmetry axis to the tangent point is $ \ 2 \ \sqrt{k} \ p \ \ = \ 10 \ p \ $ . From the coordinates of the points, we obtain
$$ (10 \ p)^2 \ \ = \ \ \left( \frac{11}{13} \ - \ 7 \right)^2 \ + \ \left( \frac{179}{13} \ - \ 13 \right)^2 \ \ = \ \ \left( \frac{11 \ - \ 91}{13} \right)^2 \ + \ \left( \frac{179 \ - \ 169}{13} \right)^2$$
$$ \Rightarrow \ \ 100 \ p^2 \ \ = \ \ \left( \frac{-80}{13} \right)^2 \ + \ \left( \frac{10}{13} \right)^2 \ \ = \ \ \frac{6500}{13^2} \ \ \Rightarrow \ \ p \ = \ \frac{\sqrt{65}}{13} \ \ \text{or} \ \ \sqrt{\frac{5}{13}} \ \ . $$
With this information in hand, we could go on to find the location of the vertex of the parabola and the equation of its directrix, but none of that was requested. The length of the latus rectum of the parabola is $$ \ 4 \ p \ = \ \frac{4 \ \sqrt{65}}{13} \ \ . $$
I am curious as to how much time and what resources were available in this "weekly test". While none of the calculations shown by the responders are terribly lengthy, providing a description of the techniques and formulas to be applied to the satisfaction of a "grader" -- particularly for a rotated conic section -- would seem to require a fair amount of writing.