$\Lambda$ is just some arbitrary indexing set. We could phrase this without an indexing set at all, as:
Let $\mathcal{A}$ be a set of sublattices of $L$. Then $$\bigcap_{X\in\mathcal{A}}X$$ (the set-theoretic intersection [of the sublattices in that set]) is also closed under meet and join. etc.
"$\{A_\lambda: \lambda\in\Lambda\}$" is the same as the $\mathcal{A}$ here, except that we've explicitly "named" the elements of $\mathcal{A}$ for easy reference, with $\Lambda$ being the set of "names" we're using.
In this specific context I think the use of an indexing set isn't helpful, but in other contexts it can substantially simplify things.
As to the definition of $[H]$, the point is that we're taking $\mathcal{A}$ to be the set of all sublattices of $L$ containing $H$. The intersection $\bigcap_{X\in\mathcal{A}}X$ is then (by the remark above) a sublattice of $L$, and it clearly contains $H$ and is a subset of every sublattice of $L$ containing $H$. So we have:
Let $H\subseteq L$ and let $\mathcal{A}$ be the set of sublattices of $L$ containing $H$. Then $$[H]:=\bigcap_{X\in\mathcal{A}}X$$ (the set-theoretic intersection of those sublattices) is the smallest sublattice of $L$ containing $H$; that is, $[H]$ is a sublattice of $L$, $H\subseteq [H]$, and whenever $J$ is a sublattice of $L$ with $H\subseteq J$ we have $[H]\subseteq J$.
An aside:
It's worth noting that any set can be "converted" into an indexed set, albeit in a rather silly way: use the set itself as the indexing set! That is, suppose I have a set $\mathcal{S}$. Let $\Lambda=\mathcal{S}$ and for $\lambda\in \Lambda$ - that is, for $\lambda\in S$ - let $S_\lambda=\lambda$. Then $\mathcal{S}=\{S_\lambda: \lambda\in \Lambda\}$.
This is admittedly really really silly, but it does tell us that we can always switch from sets to indexed sets without actually doing anything; it's just an introduction of extra language and symbols, which hopefully(!) improves the presentation. Now like I said, in this specific case I think the use of an indexing set actually just makes things messier, but it is often quite convenient.
(That's not to say that there aren't subtleties around indexed sets. For example, the axiom of choice is equivalent to "For every set $\mathcal{S}$ there is some indexing set $\Lambda$ and some 'naming surjection' $f:\Lambda\rightarrow\mathcal{S}$ - basically, $f$ tells us how $\Lambda$ indexes $\mathcal{S}$ - such that $\Lambda$ is well-orderable." But as long as we just want some indexing set, that's not a problem.)
The diagrams of $L_1$ and $L_2$ are misleading representations of lattices since they are not Hasse diagrams.
This is because in $L_1$ you have one line to much: the one from $c$ to $b$.
It is redundant because $c < f < b$, whence $c < b$ follows from transitivity.
Erase that line and it becomes obvious that the lattice is distributive (up to isomorphism, it is the lattice $\mathbf 3 \times \mathbf 2$).
Likewise in $L_2$ you have three lines to much: between $c$ and $d$; between $c$ and $e$; and between $b$ and $a$.
Again, erase those and everything becomes clear.
In the sub-lattice definition, the meaning of "same meet and join operations" is that if $a$ and $b$ are in the subset which we are testing if it is a sub-lattice, then for that set to be a sub-lattice, $a \wedge b$ and $a \vee b$ must belong to that set, where these operations are calculated in the main lattice.
So you can have a subset which is a lattice on its own, but not a sub-lattice (for example, in the lattice $L_2$ above, erase $b$, and the resulting subset is a lattice, but it is not a sub-lattice of $L_2$ because $d \wedge e = b$, which doesn't belong to the resulting subset; it is still a lattice where $d \wedge e = c$).
Best Answer
Yes, it is. This is implied in the second definition; the symbols $\land$ and $\lor$ are being used to refer to the operations in $L$.