Is (and) and (or) / (and) or (or) the same as and / or? Is there a case where it will not be? Consider the two cases:
I)
$$(a \land b) \land (a \lor b)$$
II)
$$(a \land b) \lor (a \lor b)$$
I am not familiar with the exact boolean algebra but using a truth table both come out equivalent to their inner $\land$ (I) and respectively $\lor$ (II):
a b  a∧b a∨b  I II
0 0 | 0   0 | 0 0
0 1 | 0   1 | 0 1
1 0 | 0   1 | 0 1
1 1 | 1   1 | 1 1

But what puzzles me is that this does not really fit my intuition. It feels like something different to accept either an and or an or relation than just an or (case II). Also if I want some relation to being or and and,  I can not except just an and (case I) or can I? 
It gets even better if you cascade them deeper :)
$$((a \land b) \lor (a \lor b)) \land ((a \land b) \land (a \lor b))$$
Is there anything on that in the literature?

Edit: To clarify. I am not saying that both are equivalent but if it fits your intuition that they are equal to their simpler versions 
I)
$$(a \land b) \land (a \lor b) =^? a \land b$$
II)
$$(a \land b) \lor (a \lor b) =^? a \lor b$$
 A: It seems intuitive to me because:
3) $a \lor b$ is a nescessary condition for $a \land b$ so $a \land b$ "eats" $a \lor b$ and $a \land b = (a \land b) \land $... any necessary conditions for $(a\land b)...$.  So $a \land b = (a \land b) \land (a \lor b)$.
Likewise $a \land b$ is one possible way that $a \lor b$ could be true.  So $a \land b$ gets "absorbed" by $a \lor b$.  And $a \lor b = (a \lor b) \lor $... any of the potential possibilities such that $a \lor b$ would  be true...  So $a \lor b = (a \lor b) \lor (a \land b)$. 
1) Whenever $a \land b$ is true then both $a$ and $b$ are true so $a \lor b$ is true.  So whenever $a \land b$ is true it is also the case that $a \lor b$ is true bot $a \land b$ and $a \lor b$ are true.  So $(a\land b) \land (a\lor b)$ is true.
Likewise if $a \lor b$ is true it is possible that both $a$ and $b$ are true.  So $(a\lor b) \lor (a\land b)$.
Basically $a \lor b$ must be true in order for $a \land b$ is true.  So we have $P = a \land b$ and $Q = a \lor b$ then $P \implies Q$ is a given.  So i) $P \land Q \iff P$ should be intuitive.  $P \implies Q$ can only be false if $Q$ is false and $P$ is true. So $Q \lor P \implies Q \implies Q \lor P$.  So ii) $Q \lor P \iff Q$.  
i) is $(a\land b) \land (a\lor b) = (a\land b)$
and ii) is $(a \land b) \lor (a \lor b) = (a \lor b)$.
2) It helps if you think in terms of sets. 
Bear with me.  
Suppose $a$ is the statement "Kim Kardashian is the president of the united states".  Then $a$ is true means "Of all the possible universes that can possibly be, we live in one of the universes in which Kim Kardashian is the president of the united states".  And if $b$ is the statement "cows eat grass" then $b$ is true means "Of all the possible universes that can possibly be, we live in one or the universes in which cows eat grass."
Keep bearing with me.
If $x$ is a statement let $X$ (capital $x$) = $\{$ all universe where $x$ is true $\}$.
And let's say $u = $ the one single universe in which we live in.
Now... SHOWTIME!:
$x \land y$ means $u \in X \cap Y$.
$x \lor y$ means $u \in X \cup Y$.
$x \implies y$ means $Y \subset X$
So $(a \land b) \land (a \lor b)$ means $u \in (A \cap B) \cap (A \cup B) = A\cap B$ means $a \land b$.
And $(a \land b) \lor (a \lor b)$ means $u \in (A \cap B) \cup (A \cup B) = A \cup B$ means $a\lor b$.
---- old -----
Look at your table again:
a b  a∧b a∨b   I   II
0 0 | 0   0 |   0   0
0 1 | 0   1 | **0 * 1**
1 0 | 0   1 | **0 * 1**
1 1 | 1   1 |   1   1

They are not equivalent.  
A: $$\begin{align}I.\;\;(a \land b) \land (a \lor b) & \equiv (a\land b \land a) \lor (a \land b \land b)\tag{Distributive Law} \\ \\ 
& \equiv (a \land b)\lor (a\land b) \tag{Simplification}\\ \\ 
& \equiv (a\land b)\tag{Simplification}\end{align}$$

$$\begin{align} II.\;\;(a \land b) \lor (a \lor b) &\equiv (a\lor a\lor b)\land (b\lor a \lor b)\tag{Distributive Law}\\ \\
&\equiv (a\lor b) \land (a\lor b)\tag{Simplification}\\ \\ &\equiv (a\lor b)\tag{Simplilfication}\end{align}$$

Compare the truth tables for $(I)\;\;a\land b,\;\;$ vs. $\;\;(II)\;\; a\lor b$. 
A: When we write "and/or" in natural language, the point is to make the meaning unambiguous for a reader who interprets "or" as exclusive "or". Such a reader would be working with this truth table:
a  b   a and b    a xor b    (a and b) xor (a xor b)
-----------------------------------------------------
0  0      0          0                  0
0  1      0          1                  1
1  0      0          1                  1 
1  1      1          0                  1

So we have managed to express the truth table of $\lor$ using only "and" and "exclusive or".
If the "or" you're working with is already an inclusive one like $\lor$, then there is no point in saying "and/or" instead of simply "or" -- except perhaps to explicitly remind the reader that both of the things can be true.
A: 
Edit: To clarify. I am not saying that both are equivalent but if it fits your intuition that they are equal to their simpler versions
I) $(a \land b) \land (a \lor b) \overset?= a \land b$
II) $(a \land b) \lor (a \lor b) \overset?= a \lor b$

Yes.  Yes they are.

But what puzzles me is that this does not really fit my intuition.

Perhaps your intuition is that $\lor$ is exclusive.   In fact, it is inclusive.   $a\lor b$ is "It may be $a$ or $b$ (or both)".   That is, it includes he possibiliy of both.   We don't actually need the parenthical "(or both)" except to be really clear about this.   Sometimes in casual language we say "and/or" to indicate this inclusivity as though it were an exception, but in logic statements, inclusivity is the default for $\lor$.
So $(a \land b) \land (a \lor b)$ is "It is $a$, it is $b$, and it also may be $a$, $b$ or both."   Which is simply "It is $a$ and it is $b$".
So $(a \land b) \lor (a \lor b)$ is "It is either both $a$ and $b$, or it may be $a$ or $b$ (or both)."   Which is simply "It is $a$ or it is $b$ (or both)".
A: They are not equivalent, you can use truth table to see it. What we have is: $(a∧b)∧(a∨b)\to (a∧b)∨(a∨b)$.
A: What is behind these are the equivalence principles of Absorption:
Absorption
$a \land (a \lor b) = a$
$a \lor (a \land b) = a$
And to understand these more intuitively:
For the first one: If you claim that $a \land (a \lor b)$ is true, then you are claiming that both $a$ and $a \lor b$ are true. But of course, ince $a$ is true, $a \lor b$ is automatically true as well, so the latter statement does not add any more information to $a$ itself, and hence it is equivalent to just $a$
More general: a conjunction of two statements, one of which is stronger than (i.e. implies ) the other, is equivalent to just the stronger claim. Which is what you see with $(a\lor b) \lor (a \land b)$ as well: $a \land b$ is stronger than $a \lor b$, and hence it is equivalent to just $a \land b$
For  the second one: if I tell you that either $a$ is true of $a \land b$ is true, then in either case $a$ is true, and since the former claim says exactly that, you can't infer anything more than that, and hence it is equivalent to $a$
More general, a disjunction of two statements, one of which is weaker than (i.e is implied by) the other, is equivalent to the weaker statement. And that you see with $(a \land b) \lor (a \lor b)$: $a \lor b$ is weaker than $a \land b$ and hence the disjunction of those two statements is just $a \lor b$.
A: And/or is a stochastic logic operator.
In general the term and/or is a stochastic distribution from the set of operators {and, or, xor}. Most common is the set {or, xor}.
And/or can not be described as a non stochastic logic operator thus there is no explicit single valued way to express it. Any such interpretation is a degenerate distribution and can explain the difficulties and variation in interpretation. Note that a binary distribution from the subset {and, or, xor} like {or, xor} or {and, or} can also be degenerate distributions. There are typically no cases of {and, xor}.
