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:

1. $M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.

2. $M$ has a definable well-ordering of the universe.

3. Every definable nonempty set in $M$ has a definable element.

4. Every definable nonempty set in $M$ has an ordinal-definable
element.

5. Every $\Pi_2$-definable nonempty set in $M$ has an
ordinal-definable element.

6. 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?](http://mathoverflow.net/a/180850/1946), 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?](http://mathoverflow.net/a/180850/1946).