Calculating power dissipated at a resistor with Thevenins and Nortons should they give the same answer?
This depends on what you mean by "at a resistor".  To the outside world, a Thevenin and Norton source with the same open-circuit voltage and short-circuit current are identical.  That's rather the point of the Thevenin and Norton theorems.  Whether you think of the black box as being a Thevenin or Norton source is merely a matter of your own convenience.

Therefore, anything you connect to such a black will operate identically whether you happened to think about it as a Thevenin or Norton source.  Some resistor connected to it will dissipate the same power regardless of your conception of how it works or its actual internal implementation.

However, if you are talking about the resistor that is part of the Thevenin or Norton source, then there is a significant difference between the two.  Let's say your source is 10 V in series with 5 &Omega;, or equivalently, it's 2 A in parallel with the 5 &Omega;.  Both produce 10 V open circuit and 2 A shorted, and are identical to the connected circuit, but the dissipation in this internal resistance is opposite between the two.

The Thevenin source has 0 current thru the resistor when open, so dissipates 0 W.  When shorted, there is 2 A thru the 5 &Omega; resistor, so dissipates 20 W.  The Norton source has 2 A thru the resistor for 20 W when open, and 0 A for 0 W thru the resistor when shorted.  In this sense, you can tell the difference between a black box Thevenin and Norton source.  The Thevenin source will dissipate no heat when open, and maximum heat when shorted.  The Norton source will dissipate maximum heat when open and no heat when shorted.  Again, to the electrical load, the two are identical.