Let $V$ be a free graded module. $\wedge V$ be the free commutative graded algebra.
$\wedge V$ = symmetric algebra $(V^{even})\otimes$ exterior algebra $(V^{odd})$
I don't understand this equation so I search the internet for more information but there is nothing related with free commutative graded algebra. Could you please explain it and give some materials for me? Then I can learn it by myself.
I would rather call your construction the "free graded-commutative algebra" on $V$ : it is not graded and commutative, it is graded-commutative, meaning that if $x$ and $y$ are pure of respective degree $n$ and $m$, then $xy=(-1)^{nm}yx$. I will more or less assume in what follows that your grading is over $\mathbb{Z}/2\mathbb{Z}$, you can adapt if it is in fact over $\mathbb{Z}$.
The word free pretty much always means the same thing : you have a category $GrComAlg$ of graded-commutative algebras, and a forgetful functor $U$ from this category to the category $Mod$ of modules, and you are looking for its left adjoint $F$. Then the free graded-commutative algebra on $V$ is $F(V)$.
Now in concrete terms, you want the most general graded-commutative algebra generated by $V$. This means that it should satisfy that for any graded-commutative algebra $A$ and any graded module morphism $f:V\rightarrow A$ there should be a unique graded algebra morphism $\tilde{f}: \Lambda V\rightarrow A$ extending $f$ (I say "extend" like $V$ is included in $\Lambda V$, but a priori you only have a canonical morphism $\varphi:V\to \Lambda V$).
Write $V =V_0\oplus V_1$ where $V_0$ is the even component and $V_1$ is the odd component, and similarly $A =A_0\oplus A_1$. Giving $f$ is equivalent to giving $f_0:V_0\rightarrow A_0$ and $f_1:V_1\rightarrow A_1$. By the universal property of symmetric algebras, $f_0$ extends to a unique algebra morphism $g_0: Sym(V_0)\rightarrow A_0$ because $A_0$ is commutative. On the other hand, if $2$ is invertible (which I assume is a hypothesis here, otherwise things might get messy), then $A_1$ is alternated (ie $x^2 = 0$ for all $x\in A_1$) so there is a unique algebra morphism $g_1: Ext(V_1)\rightarrow A$ extending $f_1$ (I use the non-standard notation $Ext$ because you have already used $\Lambda V$). Now define $\tilde{f}:Sym(V_0)\otimes Ext(V_1) \rightarrow A$ by $\tilde{f}(x\otimes y) = g_0(x)g_1(y)$ to build the unique extension of $f$ to your $\Lambda V$, which shows that it satisfies the universal property.
Note that writing $\Lambda V = Sym(V_0)\otimes Ext(V_1)$ is a little awkward because it has the expected module structure, but the elements of $Sym(V_0)$ and of $Ext(V_1)$ commute in $\Lambda V$.
I don't really know any specific material for this construction, but it is pretty straightforward if you're familiar with graded algebras (I made it up on the spot, so I hope there is no mistake).