How do we know the rank is 1? Here's matrix A:
$$
\begin{pmatrix}
2 & 1 \\
-4 & -2 \\
-2 & -1 \end{pmatrix}
$$
Apparently, we can determine that matrix A is of rank 1 by noting that.
$$
\det
\begin{pmatrix}
2 & 1 \\
-4 & -2 \end{pmatrix}
=
\det
\begin{pmatrix}
2 & 1 \\
-2 & -1 \end{pmatrix}
=
\det
\begin{pmatrix}
-4 & -2 \\
-2 & -1 \end{pmatrix}
=0
$$
Why does this statement guarantee that A is of rank 1? What happens, for instance, if 
$$
\det
\begin{pmatrix}
2 & 1 \\
-4 & -2 \end{pmatrix}
= 1?
$$
Also, is it possible for a matrix to be of rank 0?
 A: What you are noting is that each row of the matrix is a constant multiple of the other rows. So Row 2 is $-2\times $ Row 1, Row 3 is $-1 \times $ Row 1. 
So the rows are linearly dependent on one row only: $R_2 = -2 R_1\neq 0;\;\;R_3 = -1R_1 \neq 0$ 
When the determinant of a matrix, or submatrix ($2\times 2$ in this case) is zero, that means that the rows are linearly dependent. By testing the determinant of each $2 \times 2$ submatrix, as you did, you can determine the exact rank by determining
If the third row of your matrix had been $(5, 3)$, but all other rows the same entries as now, then the matrix would have rank $2$ (because the determinant of any submatrix containing that row would evaluate to a non-zero scalar, but the other two rows would remain linearly dependent. Your matrix would then have rank 2.
The zero matrix has rank 0.
If we use row reduction (Gaussian elimination), we can count the number of non-zero rows that remain and this in turn happens to be the rank of the matrix.
A: The size of this matrix is $3\times 2$ so $\mathrm{rank}A\leq \min(2,3)=2$ and since the column $C_1=2C_2\not=0$ so 
$$\mathrm{rank}A=1.$$
A: If you can find a sub-matrix with dimension $2\times2$ that has a non-zero determinant, then the rank is $2$
rank is the largest dimension you can find a non-zero determinant for...
a matrix filled with zero's has rank $0$.  
A: I would note that the second column is a multiple of the first, thus computing Gaussian elimination would leave only the top row thus the rank is the number of rows left after this reduction, i.e. 1
A: A matrix as rank 0 iff it is the 0 matrix. 
To compute the determinants here is not the best choice in this case as the first column is two times the second one 
A: Preliminaries: The rank of a matrix has many equivalent definitions, one of those is the dimension of the linear space spanned by the rows. 
A square matrix has determinant $=0$ if and only if its rows are linearly dependent. 
Answer to question 1:
Hence the equations you wrote mean that any two rows are linearly independent. Hence the dimension of the space spanned by the rows is at most $1$.
Answer to question 2:
Had one of the determinants was nonzero, there were at least $2$ linearly independent rows, hence the rank would have been at least $2$. But since the rows are pairs, the dimension is also at most $2$, hence equals $2$.
Answer to the last question: 
A matrix has rank zero if and only if it is the zero matrix (since the zero space is the only space with dimension zero...).
