Well, mathematically it should be easy to see.
U=qV, where U is potential energy, q is point charge and V is voltage (electric potential).
V = kQ/r.
Your two scenarios should be explainable through those equations. We'll say they're both <non-zero x,0,0> so we don't have to deal with multiple axes.
(-) test charge moves towards a (+), what happens to voltage and potential energy?
Well remember that voltage just has an arbitrary zero. We generate a potential just from a source charge in space at a distance.
If we decrease r, V must increase in magnitude as a result (V=kQ/r). If we look at our potential equation, U = (-q)V = k(-q)Q/r. Since we decrease r, U increases in magnitude. But it's a negative charge. We lost more potential energy.
What happens when a negative charge reaches a positive charge? Are we going to have to put in energy to separate those charges? Wouldn't that logically mean we are at less potential energy than if they were separated at a greater r?
(+) charge moving away from a (+) charge, what happens to voltage and potential energy?
This time we are INCREASING our r value, so V=kq/r shows Voltage must decrease.
U = kQq/r. Both of our charges are positive, but increasing r decreases our U.
