Cube and unit cubes

Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.

Here is a possible line of reasoning: when you paint the $3\times 3\times 3$ cube, you paint a total of $$ 9 \cdot 6 = 54 $$ faces (each face of the large cube consists of a $3\times 3$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).

On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of $$ 3 \cdot 8 = 24 $$ painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e. $$ P(\text{the painted face is from a corner}) = \frac{\text{number of successful outcomes}}{\text{total number of outcomes}} = \frac{24}{54} = \frac{4}{9}. $$

No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube?

The simple solution is that we have selected a random black face. There are $54$ black faces, $24$ of which are on corner cubes, so the chance we have a corner cube is $\frac {24}{54}$

You can also use Bayes' theorem. The prior probability of choosing a corner cube is $8 \over 27$. The probability of a corner cube showing a black face is $1 \over 2$. The overall probability of a black face is ${54 \over 162}=1/3$, so

$P({\rm corner }|{\rm black})={1/2 \over1/3} \times 8/27 =4/9$