Rigourously overcomplicating the problem: Let dy = distance boy travels north in time dt, and dx = distance girl travels east in time dt. We know that dy = 5dx from the question, hence:

dy | |__ dx

And the distance they separate dS in time dt is clearly the hypoteneuse. So we can write:

dS = sqrt(dx^2 + dy^2)

And divide through by dt:

dS/dt = sqrt( (dx/dt)^2 + (dy/dt)^2 )

Simply gives the rate of separation dS/dt as 5.1 feet per second.

Except now the walking speed of 1 m/s (3.6 km/h) and running speed of 5 m/s (18 km/h) are realistic

i am running at 5 frames per second

I'm good with intuiting 1-3 meters, and I have a decent estimate for 1 centimeter. Beyond that, about all I got (yes I know this is all very rough estimates) is "a kilogram is a couple pounds, a kilometer is a short mile, a liter is basically a quart". But I guess I don't even have a good intuition for a quart because that in my head rests at the crossroads of "4 cups or ¼ gallon".

Make metric meaningful early in life, or actively make referencing real metric measures a part of your life, if you want it to stick.

The implication that the boy is strafing while maintaining a straight vertical line to the girl is hilarious

You actually only need to know the latitude for that... except the local terrain will play a larger role anyway, unless they started very close to a pole and follow rhumb lines (in this case ♂︎ a meridian and ♀︎ a circle of latitude) as opposed to great circles, so better just ask for full coordinates.

What? The teacher does not want to talk about it? Let's find out anyway, to the best of my abilities. For now, we'll be assuming Earth is a fully walkable ellipsoid.

We don't have many data points in the question so let's extrapolate their movement into the past. There is the hint that they met 8 years earlier at the same spot, during which he'd have gone 1 262 304 000 ft or 384 750.2592 km, completing 9.617 polar circumferences of the Earth (40 007.863 km each).

Huh, that's not a whole number. In some languages, "eight long years" might mean "a little over 8 years" so let's assume he finished exactly 10 polar circumnavigations, which took 8 years and about 116 days. Her walked distance over that time is 5x smaller, 2 polar circumnavigations' worth or 80 015.726 km. This is only exactly 2 great circles (ellipses, really) if they are polar, but we know that it's impossible to go due east from either pole. Therefore, we'll use the other option you pointed out, of her having gone at a constant bearing of 090, her path being a circle of latitude (aka a "parallel"). To end up in the same spot, she must have not-quite-circumnavigated-but-enough-for-Phileas-Fogg the Earth (aka crossed every meridian but not the equator) an integer number of times. After a simple conversion, we can construct a table of the options.

To calcuate latitude from circle-of-latitude circumference (colc), we'll be using geodetic ↔ ECEF conversion equations (except those with the perverse prime vertical radius of curvature 𝑁 of course) and their notation (simplified with 𝑦 = 0, 𝜆 = 0, ℎ = 0 to ignore longitude and elevation) with values of the WGS-84 ellipsoid. The relationship we're seeking is between colc/2𝛑 = 𝑝, circle-of-latitude radius, which is at zero longitude equal to ECEF 𝑋, and 𝜙 (latitude). See also Wikipedia on Earth radius by location but remember to skip anything with 𝑁, we're not doing that.

The geocentric radius (𝑅) is related to 𝜙 (latitude) like this but we only need the distance to axis of rotation 𝑝.

(𝑍/𝑝)(cot 𝜙) = (1 − 𝑒²) → (𝑏²/𝑎²)(𝑍/𝑝) = 1/(cot 𝜙) = tan 𝜙 → 𝜙 = atan((𝑏²/𝑎²)(𝑍/𝑝))
(using 𝑒² = 1 −  𝑏²/𝑎²)

Since sin² 𝛂 = 1 − cos² 𝛂 and we can normalize 𝑍 and 𝑝 to the unit circle with ellipsoid radii 𝑏 and 𝑎 respectively:
𝑍²/𝑏² = (𝑍/𝑏)² = 1 − (𝑝/𝑎)² = 1 − 𝑝²/𝑎², therefore 𝑍 = √(𝑏²(1 − 𝑝²/𝑎²)).

All in all, 𝑝 → 𝜙 conversion is:
𝜙 = atan((𝑏²/𝑎²)(√(𝑏²(1 − 𝑝²/𝑎²))/𝑝))

(Presumably, this could be simpified further but I can just put this into a calculator so idc)
Per WGS-84:
𝑎 = 6378.137 km
𝑏 = 6356.752 km

Here are the results. Finding appropriate meeting locations at some of the 25+ possible latitudes on either hemisphere is left as an exercise to the reader. Also note that "rainy days" don't occur in some places, which is why I didn't bother adding more rows after I got within 500 km of the pole.

Rows where the number of not-quite-circumnavigations is divisible by 2, 5, or 10 are especially interesting because then the couple would meet 3, 6 and 11 times over the 8.32-year relationship, respectively, rather than just twice.

(Fun fact: another set of latitude circles whose sizes are an inverse-integer sequence is a lesser-known solution to the famous cheeky bear color problem (the well-known solution is a 1-radian arc around the North Pole), although obtaining the ultimate answer then fails due to the lack of bears in Antarctica)

What a convoluted way of asking the teacher to spill the beans. I like it.

It’s flat, idiot

Also, where's the topographic map of the region? How can you expect us to come up with something remotely accurate without knowing this, is the third dimension a joke to you, are we all dots of ink on a paper?

Is he due north of the starting point, or of the girl? If the former, he is traveling due north. If the latter, his 5ft/sec velocity has a 1ft/sec eastward component, and we need to calculate the northern component.