MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

Qn 1 is trivial because you said "topological spaces" rather than "compact Hausdorff spaces" (or "locally compact Hausdorff" would be okay, I guess). Simply $\lbrace 0,1\rbrace$ with the order topology and $\lbrace 0\rbrace$ will do.
If we refine to "compact Hausdorff spaces" then I take $C(X,\mathbb{R})$ and $C(Y,\mathbb{R})$, complexify, and apply GN to recover $X$ and $Y$, thus I claim that no counterexample exists.
I think that the issue stems from a confusion between the complexification of a real algebra and the underlying real algebra of a complex one. Since I can recover $C(X,\mathbb{C})$ from $C(X,\mathbb{R})$, all the information about the former is captured in the latter. However, since I can find several complex structures on the same real algebra, $C(X,\mathbb{C})_{\mathbb{R}}$ does not contain all the information that is contained in $C(X,\mathbb{C})$. There is a reason why they are called forgetful functors!
So $C(X,\mathbb{R})$ is not the underlying real algebra of $C(X,\mathbb{C})$, but $C(X,\mathbb{R})$ C(X,\mathbb{C})$is the complexification of$C(X,\mathbb{R})$. 1 Qn 1 is trivial because you said "topological spaces" rather than "compact Hausdorff spaces" (or "locally compact Hausdorff" would be okay, I guess). Simply$\lbrace 0,1\rbrace$with the order topology and$\lbrace 0\rbrace$will do. If we refine to "compact Hausdorff spaces" then I take$C(X,\mathbb{R})$and$C(Y,\mathbb{R})$, complexify, and apply GN to recover$X$and$Y$, thus I claim that no counterexample exists. I think that the issue stems from a confusion between the complexification of a real algebra and the underlying real algebra of a complex one. Since I can recover$C(X,\mathbb{C})$from$C(X,\mathbb{R})$, all the information about the former is captured in the latter. However, since I can find several complex structures on the same real algebra,$C(X,\mathbb{C})_{\mathbb{R}}$does not contain all the information that is contained in$C(X,\mathbb{C})$. There is a reason why they are called forgetful functors! So$C(X,\mathbb{R})$is not the underlying real algebra of$C(X,\mathbb{C})$, but$C(X,\mathbb{R})$is the complexification of$C(X,\mathbb{R})\$.