Solution:

a.
We have
$$4 f(x)+f(y)+f(z) \stackrel{(*)}{\geq} 4(f(x)+f(y+z)) \stackrel{(*)}{\geq} 16 f(x+y+z) .$$
Adding to this the cyclically permuted versions $f(x)+4 f(y)+f(z) \geq \ldots$ and $f(x)+f(y)+4 f(z) \geq \ldots$, we obtain $6(f(x)+f(y)+f(z)) \geq 48 f(x+y+z)$ and, dividing by 6, the claim.

b.
Yes, such $f, x, y, z$ exist. Consider the piecewise affine-linear function $f$ with breakpoints $\left(2^{k}, 2^{-k}\right)$ for $k \in \mathbb{Z}$, explicitly: for each $k \in \mathbb{Z}$ let $f$ be defined on the interval $\left[2^{k}, 2^{k+1}[\right.$ by
$$f(x)=-\frac{x}{2^{2 k+1}}+\frac{3}{2^{k+1}} \quad \text{ for } 2^{k} \leq x<2^{k+1} .$$
Thus $f$ is defined for all $x \in \mathbb{Q}^{+}$. For all $x$ we have $f(2 x)=\frac{1}{2} f(x)$, so (*) becomes equivalent to the convexity inequality $\frac{1}{2}(f(x)+f(y)) \geq f\left(\frac{x+y}{2}\right)$. The function $f$ is convex, hence also elastic. With $x=y=z=1$ we have
$$f(x)+f(y)+f(z)=3<9 \cdot \frac{3}{8}=9 f(x+y+z) .$$