a) $$x\sqrt{3x-y^2}+y\sqrt{3-x^2}=3$$$$x\sqrt{3-y^2}=3-y\sqrt{3-x^2}$$$$x^2(3-y^2)=9-6y\sqrt{3-x^2}+y^2(3-x^2)$$$$3x^2-x^2y^2=9-6y\sqrt{3-x^2}+3y^2-y^2x^2$$$$x^2=3-2y\sqrt{3-x^2}+y^2$$$$2y\sqrt{3-x^2}=3+y^2-x^2$$$$4y^2(3-x^2)=9+y^4+x^4+6y^2-6x^2-2x^2y^2$$$$12y^2-4y^2x^2=9+y^4+x^4+6y^2-6x^2-2x^2y^2$$$$9+(y^2+x^2)^2-6(y^2+6x^2)=0$$$$(-3+x^2+y^2)^2=0$$$$-3+x^2+y^2=0$$$$x^2=3-y^2$$So there is an infinite amount of solutions. I'll post b later

As DVDTSB posted above $x^2+y^2=3$. Thus for b we must show that $x^2+y^2=3z^2$ has no positive integer solutions. Since $n^2\equiv 0,1\text{ (mod 3)}$ for integer $n$, $3|x,y$. Thus $9|3z^2$ and $3|z$. Infinite descent.

solution for a) suppose that $y=3sin\beta$ and $x=3sin\alpha$ we have the followings: $3\sqrt{3}sin\alpha cos\beta+3\sqrt{3}sin\beta cos\alpha=3$ $\Rightarrow sin(\alpha+\beta) = \frac{\sqrt{3}}{3}$ $\because x,y\in[0,\sqrt{3}]\therefore sin\alpha,sin\beta\in[0, \frac{\sqrt{3}}{3}]$ obviously there are countless $x$and$y$ are possible. for example when$sin\alpha=sin\beta= \frac{\sqrt{3}}{6}$suppose$\alpha=\beta=\gamma$ and than just make another pair of $\alpha$and$\beta$which $\alpha+\beta=2\gamma$