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Absolutely true on astronomical scales.

An unnecessary complication if you're dropping a brick out of a window.

It's funny how much of physics we do assuming a flat earth.

If you did it "properly" you would calculate the orbit of the brick (assuming earth was a point mass), then find the intersection between that orbit and earth's surface. But for small speeds and distances you can just assume g points down as it would in a flat earth

To do it properly you have to account for the gravitational pull of the building.

Your physics Prof is correct of course, and so is GP. "Standard" values for g exist for these bodies, but it also varies everywhere.

This is correct, gravitational constants are a good approximation/simplification since the mass of solar bodies is usually orders of magnitude greater than the other bodies in the problem, and displacement over the course of the problem is usually orders of magnitude smaller than absolute distance between them.

In other words, we assume spherical cows until that approximation no longer works.