1. no. {1,{1,2}} is a set which has 2 elements: 1, and {1,2}. its power set is:

{Ø, {1}, {{1,2}}, {1, {1,2}} }

the set {Ø,{1,2}} is not any of these 4 elements. <--this is a set with 2 elements and the only element of P({1,{1,2}}) that has 2 elements is {1,{1,2}}, and 1 is not an element of {Ø,{1,2}}, so these two sets can't be equal.

2. it depends on your viewpoint.

strictly speaking, R^{2}is not a subset of R^{3}, since a (non-empty) subset of R^{3}consists of vectors (x,y,z) all of which have 3 coordinates. elements of R^{2}have only two coordinates (a type-mismatch).

however, there are subspaces of R^{3}isomorphic (as vector spaces, and thus as sets) to R^{2}:

one can verify that (x,y) -->(x,y,0) defines an injective map of R^{2}into R^{3}, the image of which behaves in almost every respect "just like R^{2}".

in various algebra fields, one often replaces "equal" to "equal up to isomorphism" (which is an equivalence relation, albeit one usually over a proper CLASS, not a set), because "isomorphic objects" behave the same as far as algebra is concerned.

one CAN take the same view with sets: regarding the set {a,b,c} as "essentially the same" as {a,b,d} (they both have 3 elements)...in this context a set-isomorphism is just a bijection.

BUT...obviously, this can lead to problems if one is looking at P({a,b,c,d}), for example.

this problem is a bit worse than it might appear: consider the following distinct sets:

A = {Ø, {Ø}, {Ø,{Ø}}, {Ø,{Ø,{Ø}}},.....} if we call Ø = 0, and {0} = 1, and {0,1} = 2, this can be seen as one way to construct natural numbers.

B = {Ø, {Ø}, {{Ø}}, {{{Ø}}}, .......} this starts out the same, but now 2 = {1}, 3 = {2}, etc. this is an alternate way to construct natural numbers.

but shouldn't the natural numbers be one set or the other? which is it?

another way would be to let natural numbers be "ur-elements", things which are like "atoms" that can inhabit sets, but are not sets themselves. this is probably the way sets were first thought of (later, it was decided this was "excess baggage"). the problem with doing this, however, is that it lets us do mathematics perfectly fine, but leaves us wondering what {Ted,Bob,Alice} might mean (i'm not an expert, but i believe that logicists address this by making such things part of a "background alphabet", whatever THAT means. explain one term, you get another...i find myself wondering if we even ever know what we are talking about at all...but that's another can of worms).

short answer to #2: if you're talking to a set-theorist, it's false. if you're talking to a topologist (or algebraist) it's true (up to some isomorphism, of course).