(b) is plain wrong. In fact, the opposite is true for any hermitian operator, as it is just the well-known fact that the variance of (the eigenvalues of) $W$, $\mathrm{Var}(W)=\mathrm{tr}(W^2)-\mathrm{tr}(W)^2$, is non-negative. The incorrectness of (b) for witnesses can also be easily checked by e.g. taking $W$ the SWAP operator.

(a) holds since $\mathrm{tr}(W)$ is the value the witness takes on the maximally mixed state. Since there is an open ball of separable states around the maximally mixed state, this implies that we can add $\varepsilon$ of any hermitian operator $H$ the maximally mixed state and $\mathrm{tr}((1\!\!1+\varepsilon H)W)$ is still non-negative. This is only possible if $\mathrm{tr}(W)$ is strictly positive. (To this this, you can e.g. choose $H=-W$.)