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1) If $X $ is union of two unlinked circles and $Y $ is union of two linked circles, then $\pi_1(\mathbb{R}^3-X)=\mathbb{Z} * \mathbb{Z}$, and $\pi_1(\mathbb{R}^3-Y)=\mathbb{Z} \times \mathbb{Z}$ (See Algebraic Topology-Hatcher). Can we say that $X$ and $Y$ are not homeomorphic?

2)By "cut-and-paste", we can transform one space to another. Can we say that they are homeomorphic?

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$X$ and $Y$ are homeomorphic, but their complements in $\mathbb{R}^3$ aren't. Have you reviewed the definition of homeomorphism? It makes no reference to an ambient space. – Qiaochu Yuan Sep 15 '11 at 3:51
Forget "paste" - try to think what sort of things you could cut an interval into. Or a sphere. Or anything, really. – Alon Amit Sep 15 '11 at 7:13
If things were that easy, we wouldn't have knot theory! – user641 Sep 15 '11 at 23:48
up vote 1 down vote accepted

Disjoint unions of embedded circles (that is, a collection of knots) in a 3-dimensional surrounding space are called links. The notion of equivalence used for knots and links is ambient isotopy, not homeomorphism of the circles. Here the difference in $\pi_1$ of the complements shows that there is no homeomorphism of R^3 (including orientation reversals, and not only smooth isotopies) that carries one pair of circles onto the other. Fundamental groups are a complicated way of proving that the linked circles cannot be pulled apart, since a simpler invariant, the linking number, is zero for one pair and nonzero for the other.

Cut-and-paste has several different meanings but generally the ability to cut and paste one space into a copy of the other is the opposite of the idea of homeomorphism and does not imply the spaces are topologically equivalent. Topological structure is not preserved by discontinuous operations, but some topological invariants such as the dimension of the space may be invariant under "cut and pasting".

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