Challenge 1: findall/3 Over Every Parent-Child Pair — Possible Solution
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family.pl:

parent(tom, bob).
parent(tom, liz).
parent(bob, ann).
parent(bob, pat).
parent(liz, jim).

Query:

    ?- findall(Parent-Child, parent(Parent, Child), Pairs).
    Pairs = [tom-bob, tom-liz, bob-ann, bob-pat, liz-jim].

Explanation:
The template here is the compound term Parent-Child (Prolog's "-"
operator works perfectly well as an ordinary two-argument functor for
building simple pair terms, not just for arithmetic subtraction).
findall runs parent(Parent, Child) to exhaustion via backtracking,
and for each solution it finds, it builds one Parent-Child term using
whatever Parent and Child were bound to at that point, collecting all
five into a single list in the order the facts were tried. Unlike
bagof, findall never groups these by Parent -- every pair lands in
one flat list regardless of which parent it came from.

WHY THIS WORKS AS AN ANSWER
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This uses findall's Template argument to build a compound term rather
than just collecting a single variable, demonstrating that the
template can be any term built from a query's bound variables, and
confirms findall's flat, ungrouped, all-in-one-list behavior over the
whole database.