True/false: There are $4$ linearly independent vectors $v_1,v_2,v_3,v_4 \in \mathbb{R}^3$ 
True/false: There are $4$ linearly independent vectors
  $v_1,v_2,v_3,v_4 \in \mathbb{R}^3$.

I think the statement is true.
Let's take as example (this is my own example, not of the task!!):
$$a\begin{pmatrix}
1\\ 
0\\ 
0
\end{pmatrix}+b\begin{pmatrix}
0\\ 
1\\ 
0
\end{pmatrix}+c\begin{pmatrix}
0\\ 
0\\ 
1
\end{pmatrix}+d\begin{pmatrix}
0\\ 
0\\ 
1
\end{pmatrix}=\begin{pmatrix}
0\\ 
0\\ 
0
\end{pmatrix}$$
We get that $a=b=0$
Now if $c=0$, then $d=0$, but of course we cannot know that in this case. We have to know it..?
Anyway you can see I'm not sure : /
 A: Assume $\vec v_1,\vec v_2,\vec v_3$ are linearly independent in $\mathbb{R^3}$. Then they form a basis, and we have that:
$$a\vec v_1+b\vec v_2+c\vec v_3=\vec x$$
for any $\vec x\in\mathbb{R^3}$ and for some $a,b,c\in\mathbb{R}$, and therefore includes any $\vec v_4\in\mathbb{R^3}$.
A: Vectors are linearly independent if their weighted sum (linear combination) is zero if and only if all the weights are $0$s.
This is not the case in the OP:
$$0\cdot\begin{pmatrix}
1\\ 
0\\ 
0
\end{pmatrix}+0\cdot\begin{pmatrix}
0\\ 
1\\ 
0
\end{pmatrix}+(-1)\cdot\begin{pmatrix}
0\\ 
0\\ 
1
\end{pmatrix}+(+1)\cdot\begin{pmatrix}
0\\ 
0\\ 
1
\end{pmatrix}=\begin{pmatrix}
0\\ 
0\\ 
0
\end{pmatrix}.$$
That is, the vectors above are not linearly independet.
So, the statement is false.
Above, we only showed that the example four-tuple was linearly dependent. But, it is true in general that in $R^3$ only three (or less) linearly independent vectors can be singled out. 
A: In your choice: $v_3=v_4$. Hence $v_1,v_2,v_3,v_4$ are linearly dependent:
$0=v_3-v_4$.
