a) We have $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} > 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}} \Leftrightarrow x+y+z + \sqrt{x^2+xy+yz+zx} + \sqrt{y^2+xy+yz+zx} + \sqrt{z^2+xy+yz+zx} > 2 \cdot \frac{(x+y)(y+z)(z+x)}{xy+yz+zx}$. But
$\sqrt{x^2+xy+yz+zx} > x$, $\sqrt{y^2+xy+yz+zx} > y$ and $\sqrt{z^2+xy+yz+zx} > z$, hence
$$x+y+z+\sqrt{x^2+xy+yz+zx}+\sqrt{y^2+xy+yz+zx}+\sqrt{z^2+xy+yz+zx} > 2(x+y+z).$$
It is sufficient to show that $x + y + z \ge \frac{(x+y)(y+z)(z+x)}{xy+yz+zx}$, i.e. $(x + y + z)(xy + yz + zx) \ge (x + y)(y + z)(z + x)$. After some computations, the previous inequality comes to $xyz \ge 0$, which is obviously true.

b) Fix $z = 1$. We look for $x, y > 0$, with $y = x$, such that
$$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} < k \sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}$$
i.e. $2\sqrt{x+1} + \sqrt{2x} < k\sqrt{\frac{2(x+1)^2}{x+2}}$, or $\sqrt{\frac{x+2}{2(x+1)}}(2 + \sqrt{\frac{2x}{x+1}}) < k$.
As $\sqrt{\frac{x+2}{2(x+1)}} < 1$, it is sufficient to find $x$ such that $2 + \sqrt{\frac{2x}{x+1}} < k$, i.e. $\sqrt{\frac{2x}{x+1}} < k - 2$, or, equivalently, $\frac{2x}{x+1} < (k-2)^2$. Putting $t = (k-2)^2 > 0$, the previous condition is fulfilled by any $x > 0$ if $t \ge 2$, while in case that $t < 2$, it reduces to $x < \frac{t}{2-t}$. Since $\frac{t}{2-t} > 0$, there exist positive numbers $x$, $y$, $z$ that fulfill the conditions of the statement.