$(1,1,0)$ and $(0,1,0)$ are not vectors in $\mathbb R^2$, since they have three components, so you cannot discuss their linear independence in $\mathbb R^2$.
Now, in $\mathbb R^3$, take $a\cdot (1,1,0)+b\cdot (0,1,0)=(0,0,0)\Rightarrow \begin{cases} a\cdot 1+b\cdot 0=0\\ a\cdot 1+b\cdot 1=0\\ a\cdot 0+b\cdot 0=0\end{cases}\Rightarrow a=0,\ b=0$. Therefore your vectors are linearly independent.
Working with matrices, your vectors are linearly independent if the matrix constructed with their components has rank equal to the number of vectors. The rank of your matrix is $2$, therefore the vectors are linearly independent.
Your work for part one appears to be correct, minus the typo you made (twice). The fourth coordinate of $z$ is $-7$ in the questions, and you have it as $7$ in your vector equation and augmented matrix (you clearly did the work with the other vectors though, so that's just more of a note!)
In terms of whether you need more - that would depend on your Professor, but in general I would say no. You could demonstrate that taking $a_1 = -5a_3$, $a_2 = 3a_3$ and $a_3 = a_3$ you are guaranteed to have $0$. It would look something like:
$$
1\times(-5a_3) + 1\times(3a_3) + 2\times(a_3) = -5a_3 + 3a_3 + 2a_3 = 0a_3 = 0 \\
2\times(-5a_3) + 3\times(3a_3) + 1\times(a_3) = -10a_3 + 9a_3 + a_3 = 0a_3 = 0 \\
-2\times(-5a_3) - 1\times(3a_3) - 7\times(a_3) = 10a_3 - 3a_3 - 7 a_3 = 0a_3 = 0 \\
1\times(-5a_3) + 4\times(3a_3) - 7\times(a_3) = -5a_3 + 12a_3 - 7a_3 = 0a_3 = 0
$$
It is important to note further that $z = 5x - 3y$.
For part (B) you need to find another vector space such that the Direct Sum gives you all of $\mathbb{R}^4$ We thus need $\mathbb{R}^4 = V + U$ and $V\cap U = \{0\}$
Because $V$ is linearly dependent, notably that $z = 5x - 3y$, then $V = span\{x, y\}$.
Ultimately, since you are currently dealing with a set of 2 linearly independent vectors, two more must be added such that the four together form a basis. If you add two vectors that are independent of $x$ and $y$, then by definition of linear independence you would have that $V\cap U = \{0\}$, and since you are dealing with the span of $4$ linearly independent vectors, that must form a basis for $\mathbb{R}^4$.
There are a number of ways which you can try to find these two vectors to form a spanning set. One suggestion may be to use orthogonality. This works since any non-zero, orthogonal vectors are linearly independent. Another may be to consider the following:
$$
w = \begin{bmatrix} 2 \\ 5 \\ -3 \\ 1 \end{bmatrix}
$$
This vector is constructed simply by taking a linear combination $x + y$ and then differing the 4th component from what was expected.
Thus any combination of the three can be written:
$$
z* = \begin{bmatrix} a + b + 2c \\ 2a + 3b + 5c \\ -(2a + b + 3c) \\ a + 4b + c \end{bmatrix}
$$
Now, consider:
$$
v = \begin{bmatrix} 2 \\ 6 \\ 2 \\ 1 \end{bmatrix}
$$
This matrix was derived simply by taking linear combinations of the three above, and making the system unsolvable. Specifically note that:
$$
a + b + 2c = 2 \implies a = 2 - b - 2c \\
2(2-b-2c) + 3b + 5c = 6 \implies b = 2 - c \\
\implies -(2(2-b-2c) + (2 - c) + 3c) = -2
$$
So by taking $<2, 6, X, Y>$ where $x \neq -2$ we would arrive at an insolvable system, and thus they must be linearly independent.
So then taking $U = span\{w, v\}$ you should have that they are a basis for $\mathbb{R}^4$.
Best Answer
First, note that augmenting with $0$'s is superfluous, as row operations won't change a zero-column.
It would be useful to find the reduced row-echelon form of $A$. In this case $$ \DeclareMathOperator{rref}{rref}\rref A= \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] $$ This tells us that any scalars $a_1,a_2,a_3$ satisfying $$ a_1 u+a_2 v+a_3 w=\vec 0 $$ must also satisfy $$ a_1=a_2=a_3=0 $$ This is exactly the statement that $u,v,w$ are linearly independent!