Is this set a subspace of $\mathbb{R}^4$? 
is this a subspace of R4?
1st criterion is fulfilled, because
(85,-58,11,0) is element of L5
But I dont know how to proof the 2nd and the 3rd
2nd says
λ element of R
a element of L5
a*λ element of L5
the 3rd says
a,b elements of L5 
a+b element of L5
edit:
all other criterions aside
would the 3rd aka a+b criterion be fulfilled?
a+b=(85,-58,11,0)+(85,-58,11,0)+(t1+t2)(-42,32,-7,1)
how to check if that term above is in L5?
 A: If this was a subspace it would have to contain the zero vector. But this vector is not in $L_5$. You see this when you try to solve:
$$\begin{align}
85 - 42t &= 0 \\
-58 + 32t &= 0 \\
&\vdots
\end{align}
$$
Since there is not value of $t$ satisfying all these equations, then zero vector is not in $L_5$.
Edit: Given the discussion in the comments below, it might  be that your definition of subspace doesn't directly require that the $0$ vector be an element of the space. 
Let us say that the definition of a subspace $W$ of $V$ is simply that $W$ is a subset such that 


*

*For $\lambda \in \mathbb{R}$ and $w\in W$, then $\lambda w \in W$

*For $v,w\in W$, $v + w \in W$.


Now take any vector $v \in W$. Then $-1\cdot v = -v \in W$ by 1. Then $0 = v + (-v) \in W$ by 2. So indeed a subspace must contain the $0$ vector. As Jyrki points out in the comments below, you could also note that $0 \mathbb{R}$ so for any vector $v$ you get the zero vector by $0v$.
That said, in your concrete example, note that 
$$
\pmatrix{85 \\ -58 \\ 11 \\ 0} \in L_5.
$$
If $L_5$ was a subspace then
$$
2\pmatrix{85 \\ -58 \\ 11 \\ 0} = \pmatrix{170 \\ -116 \\ 22 \\ 0}
$$
would be an element in $L_5$. But again, when you try to solve
$$\begin{align}
85 - 42t &= 170 \\
-58 + 32t &= -116 \\
&\vdots
\end{align}
$$
you find no solution.
