a) Wlog, we may assume that $ C$ is infinite and unbounded from above. Assume, for a contradiction, that there is no triples $ a\in A,b \in B, c\in C$ which are the sides of a triangle. Let $ a \in A$ and $ b \in B$. We may assume that $ a<b$. Then $ ]b,a+b[ \cap C = \emptyset$, so that $ b+ \frac a 2$ belongs to $ A$ or $ B$. Now assume that for some integer $ n>1$, we have $ ]b,b+n\frac {a}{2} [ \cap C = \emptyset$. Then $ b+(n-1)\frac {a}{2}$ belongs to $ A$ or $ B$. - If $ b+(n-1)\frac {a}{2} \in A$ then $ ]b+(n-1)\frac {a}{2}, b+b+(n-1)\frac {a}{2}[ \cap C = \emptyset$ otherwise there would exist $ c \in C$ such that $ b+(n-1)\frac {a}{2}<c<2b+(n-1)\frac {a}{2}$ so that $ b,c,b+(n-1)\frac {a}{2}$ would be the sides of a triangle, one in each sets. A contradiction. Therefore $ 2b+(n-1)\frac {a}{2} > b+(n+1)\frac {a}{2}$ (since $ b>a$) and $ ]b,b+n\frac {a}{2} [ \cap C = \emptyset$ and $ ]b+(n-1)\frac {a}{2}, b+b+(n-1)\frac {a}{2}[ \cap C = \emptyset$, thus $ ]b, b+(n+1)\frac {a}{2}[ \cap C = \emptyset$. - If $ b+(n-1)\frac {a}{2} \in B$ then $ ]b+(n-1)\frac {a}{2} ;b+(n-1)\frac {a}{2} +a[ \cap C = \emptyset$ otherwise there would be $ c \in C$ such that $ b+(n-1)\frac {a}{2} < c < b+(n+1)\frac {a}{2}$ which forces $ a,b+(n-1)\frac {a}{2},c$ to be the sides of a triangle, one in each of the three sets. A contradiction. Thus, in any case, we have $ ]b,b + (n+1)\frac a 2 [ \cap C = \emptyset$. By induction, we deduce that $ ]b, +\infty [ \cap C = \emptyset$, which contradicts that $ C$ is not bounded from above. b) Let $ A,B$ form any partition of the algebric real numbers, and $ C$ be the set of transcendental real numbers. It is known that if $ a,b$ are algebric numbers, then so are $ a^2,b^2,a^2+b^2,a^2-b^2, \sqrt {a^2+b^2}, \sqrt {a^2-b^2}, \sqrt{b^2-a^2}$ (if they exist). Then, if $ a \in A, b\in B, c \in C$ are the sides of a right triangle then $ c = \sqrt {a^2+b^2}$ or $ c= \sqrt {a^2-b^2}$ or $ c = \sqrt{b^2-a^2}$. In any case, $ c$ cannot be transcendental. A contradiction. Pierre.

b) just divide all positive reals into three disjoint groups as shown: $ A=\mathbb{N}$ $ B=\{y|\text{all positive reals such that}$ $ y^2\in Q$ $ \text{and y is not integer}\}$ $ C=\{z|\text{all reals such that}$ $ y^2$ $ \text{is not rational}\}$ so it is irrational, rational\integer and integer numbers.

a) Assume the contrary. Let $ a\in A,b\in B$ and WLOG $ a < b$. Note that $ (b,a + b)\cap C=\emptyset$. Hence for each $ x\in (b,a + b)$ we must have $ x\in A$ or $ x\in B$. WLOG $ x\in B$ and again we must have $ (b,a + x)\subset (a,a + x)\cap C=\emptyset$. Continuing this process an infinite number of times we conclude that there is no $ x > b$ such that $ x\in C$. Let $ c\in C$. Then $ c < b$ and $ (\min\{a,c\},a + c)\cap B=\emptyset$. Thus performing the same operation for $ a,c$ we conclude that there is no $ x > \max\{a,c\}$ such that $ x\in B$, contradiction. b) Define $ A,B,C$ as follows: \begin{align*}A & = \{1\} \\ B & = \{\sqrt n\mid n\in\mathbb N - A\} \\ C & = \mathbb R^ + - (A\cup B).\end{align*} It follows that in this case there is no right triangle with the property stated.

a)To say easily, numbers from $A$ be colored blue, numbers from $B$ be colored red and numbers from $C$ be colored black. Let there doesn't exist such a,b and c, which colors are different and form triangle. We know that, there exists at least one blue,at least one red and at least one black number, so randomly take 3 different colored number a,b and c. Without restriction of generality let $a<b<c$. Let's discuss such intervals : $[b+ak;b+a(k+1))$, where $k\in(\mathbb{N}\cup{0})$. Notice that fact$1$: We can't have at least one black and at least one red number simultaneously in some such interval. Let there exists such red $b_0$ and black $c_0$ and for some $t$, $b_0,c_0\in{[b+at;b+a(t+1))}$ . Take blue a, red $b_0$ and black $c_0$. $a<b_0,c_0.$ $|b_0-c_0|<a$, because of both of $b_0$ and $c_0$ are in $[b+at;b+a(t+1))$. So all of them are colored with different color and form triangle. This is contradiction , so this fact is right. Fact$2$: We can't have at least one black and at least one blue number simultaneously in some such interval. The proof is analogical, but in process we use that $a<b$. From the first and the second fact, we can say that, if there exists at least one black number at such interval $[b+ak;b+a(k+1))$, the whole interval is complected with black numbers.It mean's that every number in this interval is black. Name such intervals "Kurdi". We know that we have black numbers(at least one - $c$) after $b$. Let's discuss first "Kurdi" interval, let it be $[b+at_0;b+a(t_0+1))$. Because of that interval is the minimal "Kurdi" interval, we can say that any number from $(0;b+at_0)$ is not black, it's blue or red. So take any $b+at_0-\varepsilon$, where $\varepsilon\in(0;a))$. This number is colored blue or red,for black number take $b+at_0$ and for the blue or red color(different from $(b+at_0-\varepsilon)$'s color) take $a$ or $b$ (situation will show which is necessary). This is simple to show that this three numbers form triangle. This is condraction. So there exists such $a,b$ and $c$, which colors are different and form triangle. b) The answer will be "NO", example coming soon.

a) is same with other solutions, just take 1 of A,B,C unbounded from above and get contradiction. b) $A=\{n| n\in \mathbb{N}\}$ $B=\{\sqrt{m}| m \in \mathbb{N}\}$ where $\sqrt{m}$ is not integer. And $C$ is set of remaining numbers. And result follows immediately.