ag.algebraic geometry - When is the canonical divisor of an algebraic surface smooth?

Any smooth projective surface with nonempty $K_X$ is obtained by blowing up finitely many points on its unique minimal model. From the formula $K_X=f^*K_Y+E$ for the blowup, you see that the exceptional divisors of the blowup are always in the base locus of $|K_X|$. Thus, the problem is reduced to the minimal model.

Then you have to go through the classification of the minimal models of surfaces, that's been known for a hundred years now. (Using a book such as van de Ven "Complex surfaces", or Shafarevich et al, or Beauville...)

For a minimal surface $Y$ Kodaira dimension 0 for example, $12K_Y=0$. So either $K_Yne 0$ and then $|K_X|=emptyset$, or $K_Y=0$ and then any divisor in $|K_X|$ is $sum a_i E_i$, where $E_i$ are the exceptional divisors of the blowups.

For a minimal surface $Y$ of Kodaira dimension 2, the question is still somewhat tricky. If looking at higher multiples $|mK_Y|$ suffices, then by a well known theorem (Bombierri? certainly I. Reider gave a very nice proof), $|5K_Y|$ is free, so a general element is smooth (in characteristic 0). For $|K_Y|$ I don't think the answer is known but why not search mathscinet.

Finally, for Kodaira dimension 1, an elliptic surface $pi:Yto C$, there is a well-known Kodaira's formula for the canonical class $K_Y=pi^*K_C + R$ with explicit rational coefficients in $R$. I'd play with that. Again, for higher multiples I think $|12K_Y|$ works.

Of course, to your example of a hypersurface in $mathbb P^3$ you can add the case of complete intersections, and other surfaces for which $K_X$ is either zero or $pm K_X$ is very ample.