Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member?
The following theorem seems to express how the various definability witness properties are connected with each other and with $V=\text{HOD}$.
Theorem. The following are equivalent in any model $M$ of ZF:
$M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.
$M$ has a definable well-ordering of the universe.
Every definable nonempty set in $M$ has a definable element.
Every definable nonempty set in $M$ has an ordinal-definable element.
Every $\Pi_2$-definable nonempty set in $M$ has an ordinal-definable element.
Every ordinal-definable nonempty set in $M$ has an ordinal-definable element.
Proof. ($1\to 2$) The usual HOD order is a definable well-ordering of the universe.
($2\to 3$) Select the least element with respect to the definable order, as in Bjorn's answer.
($3\to 4$) Immediate.
($4\to 5$) Immediate.
($4\to 1$) If $M$ thinks there is a non-OD set, then the set $A$ of all non-OD sets in $M$ of minimal rank is a definable nonempty set in $M$ with no ordinal-definable elements.
($5\to 1$) The stronger implication has now undergone a few improvements, so let me discuss it. I had proposed considering as above the set $A$ of all minimal-rank non-OD sets, which is definable and nonempty in any model of $V\neq\text{HOD}$, but has no ordinal-definable elements. I had guessed that $\Sigma_5$ would be sufficient to define $A$. In the comments, François refined this, arguing that this set was actually $\Sigma_3$-definable and indeed $\Delta_3$-definable. Using his idea, I was able to push this down to show that $A$ is $\Sigma_2\wedge\Pi_2$ definable, by the properties: $A$ is not empty; all elements of $A$ have the same rank; every element of $A$ is not in OD; every set of rank less than an element of $A$ is in OD; every set not in $A$, but of the same rank as an element of $A$, is in OD. Each of these properties is either $\Sigma_2$ or $\Pi_2$, making the set $A$ to be $\Sigma_2\wedge\Pi_2$-definable. Specifically, the first two requirements are $\Sigma_2$, being witnessed in a rank-initial segment of the universe; the third is $\Pi_2$; the fourth and fifth are both $\Sigma_2$, since they are true just in case there is a large $V_\theta$ which believes them to be true. I also noted that $A$ is not provably $\Sigma_2$-definable.
Meanwhile, over at my question Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable element?, Emil made a suggestion leading to the observation that if $V\neq\text{HOD}$, then there is a $\Pi_2$-definable set with no ordinal-definable elements. The set is simply $U=A\times V_\theta$, where $A$ is as above and $\theta$ is least such that $V_\theta$ thinks $A$ is the set of minimal-rank non-OD sets. So I refer the reader to theorem 2 in that answer, which provides the content of the implication ($5\to 1$).
($1\to 6$) Immediate, since under statement $1$, every set in $M$ is ordinal-definable in $M$.
($6\to 4$) Immediate. QED
Conclusion. Thus, case (1) of the question occurs in exactly the models of $V=\text{HOD}$ that are not pointwise definable. There are such models, if ZFC is consistent, since one may take any uncountable model of $\text{ZFC}+V=\text{HOD}$.
Meanwhile, case (2) of the question — ignoring the issue of real parameters — does not occur at all, since if a set has sets that are not ordinal-definable, then it will have a definable set with no ordinal-definable members, namely, the set of all non-OD sets of minimal rank, as in the implication of statement 4 to statement 1.
Update. I edited to the improved statement 5, which we've now got down to the case of mere $\Pi_2$-definability, using the answer to my question Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable element?.
Update. This answer and those of the related questions have known grown into the following paper:
F. G. Dorais and J. D. Hamkins, When does every definable nonempty set have a definable element? (arχiv:1706.07285)
Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$.Read more at the blog post.
Yes, fix a definable relation $\le_L$ that well-orders all of $L$.
If $V=L$ then every definable nonempty set $A$ has a definable member $a$, namely:
$a :=$ the $\le_L$-least element of $A$.
The answer in case 1. is also yes. In fact, a stronger assertion is true: there exist models of set theory in which every set is definable without parameters. Such models are called pointwise definable, and (as a first observation) are necessarily countable. A collection of results surrounding pointwise definable models of ZFC and GBC (in which every set and every class are definable without parameters) are presented in "Pointwise Definable Models of Set Theory," joint work by Joel Hamkins, David Linesky and me. Here's a link to Joel's blog post on the paper, which gives an overview & link to the paper itself: http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/