Nice problem! a) Take $a=b=c=t$, where $t$ is a root of $t^2 + t = 1$. (Probably there is an example with $a$, $b$, $c$ distinct, but lazy to look for such.) b) Observe that $c + ab = 1-a-b + ab = (1-a)(1-b)$ and that $a$, $b$, $c$ are rational if and only if $x=1-a$, $y=1-b$, $z=1-c$ (with $x+y+z=2$) are such. But now we have obtained that $xy$, $yz$ and $zx$ are rational and non-zero. Hence $x^2 = (xy) \cdot (xz) / (yz)$ and similarly $y^2$ and $z^2$ are rational, so it remains to show that if $m,n,p$ non-zero rationals are such that $\sqrt{m} + \sqrt{n} + \sqrt{p}$ is rational (in this case equal to $2$), then $\sqrt{m}$, $\sqrt{n}$, $\sqrt{p}$ are also rational. If $\sqrt{m} + \sqrt{n} + \sqrt{p} = w \neq 0$, then $m+n+2\sqrt{mn} = w^2 - 2w\sqrt{p} + p$, i.e. $2\sqrt{mn} = A - 2w\sqrt{p}$, where $A = w^2 + p - m - n$. Hence $4mn = A^2 + 4w^2p - 4w\sqrt{p}$. Since $w \neq 0$, we obtain that $ \sqrt{p} = \frac{A^2 + 4w^2p - 4mn}{4w}$ is rational, hence $\sqrt{p}$ is rational. Analogously we argue for $\sqrt{m}$, $\sqrt{n}$ and we are done.

a) Let $a$ is root $a^3-a=1$ ( easy to prove, that $a$ and $a^2$ is irrational),$ b=-a, c=a^2$ Then $a+bc=a-a^3=-1, b+ac=-a+a^3=1,c+ab=a^2-a^2=0$ b) $a+bc=a(a+b+c)+bc=(a+b)(a+c)$ is non-zero rational Same way $(b+a)(b+c), (c+a)(c+b)$ are non-zero rationals, so $ \frac{(a+b)(a+c)*(a+b)(b+c)}{(c+a)(c+b)}=(a+b)^2$ is rational $ 2=2(a+b+c)=(a+b) ( 1+\frac{(a+b)(b+c)}{(a+b)^2}+\frac{(a+b)(c+a)}{(a+b)^2})$ so $a+b$ is rational and so $c=1-(a+b)$ is rational Same way for other variables.