a) The triple $(n, n+1, n+3)$, $n \in \mathbb{Z}$, is artistic iff $\frac{3n^2+8n+3}{3n+4} = n + \frac{4n+3}{3n+4}$ is an integer. It follows that $3n+4$ divides $4(3n+4)-3(4n+3) = 7$, whence we obtain $n \in \{-1, 1\}$, therefore the solutions are $(-1, 0, 2)$ and $(1, 2, 4)$.

b) Since $x + y + z$ and $\frac{xy+yz+zx}{x+y+z}$ are both integers, $\frac{x^2+y^2+z^2}{x+y+z} = (x+y+z) - 2 \cdot \frac{xy+yz+zx}{x+y+z}$ and $\frac{x^2y^2+y^2z^2+z^2x^2}{x+y+z} = (xy+yz+zx) \cdot \frac{xy+yz+zx}{x+y+z} - 2xyz$ are also integers. Thus, $\frac{x^4+y^4+z^4}{x+y+z} = (x^2+y^2+z^2) \frac{x^2+y^2+z^2}{x+y+z} - 2 \cdot \frac{x^2y^2+y^2z^2+z^2x^2}{x+y+z}$ is an integer.