Explanation to Linear Independency Please take a look at the rref(A) matrix:
$\begin{bmatrix}
1&a&0&b&d&0\\
0&0&1&c&e&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0\\
0&0&0&0&0&0
\end{bmatrix}$
The question is asking me to explain why the first three rows of the matrix in Q3 are linearly independent. I could only generate this answer:

The 1st and 2nd rows can not make the 3rd row because their last
  component is 0 and so any Similarly with the 3rd component the 2nd row
  (can't be a linear combination of rows 1 and3.) I don't think that's a
  good answer. It seems like I'm pointing out the fact that they are
  linearly independent more than "explaining" the reasons behind it.

 A: If you want a 'formal' way to prove that the third row is independent to the first two, you can do so by contradiction:

Suppose that $r_3$ is a linear combination of $r_1$ and $r_2$. Then for some $\alpha, \beta$ with not both equal to $0$, we have that $$(\alpha,\alpha a, 0, \alpha b, \alpha d, 0) + (0,0,\beta,\beta c, \beta e, 0) = \alpha(r_1) + \beta(r_2) = r_3 = (0,0,0,0,0,1)$$
  Then $0 + 0 = 1$, a contradiction. Hence $r_3$ cannot be expressed as a nontrivial linear combination of $r_1$ and $r_2$.

That essentially captures the idea of what you wrote. You can write something similar to show that $r_2$ is not a linear combination of $r_1$, and you're done.
A: I would express it a little differently, and I think that what I'm about to write is kind of the "point" of the reduced form you have the matrix in.  
Let $v_1,v_2,v_3$ be the first three rows of your matrix, and suppose 
$$
a_1v_1 + a_2v_2 + a_3v_3 = 0
$$
We know the first coordinate of the sum is $a_1$, the third is $a_2$, and the sixth is $a_3$.  This shows that all $a_i$ are zero, which shows the rows are linearly independent ( ie. there is no non-trivial linear combination of them that equals zero )
