Determine linear independence for 4.1.4.

To determine if the vectors $x_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$, $x_2 = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$, $x_3 = \begin{bmatrix} 1 \\ 2 \\ 1 \\ 2 \end{bmatrix}$, and $x_4 = \begin{bmatrix} 3 \\ 5 \\ 5 \\ 7 \end{bmatrix}$ are linearly independent, we form a matrix $M$ with these vectors as columns and calculate its determinant.

$$M = \begin{bmatrix} 1 & 1 & 1 & 3 \\ 1 & 2 & 2 & 5 \\ 1 & 3 & 1 & 5 \\ 1 & 4 & 2 & 7 \end{bmatrix}$$

Perform row operations to simplify the determinant calculation:
$R_2 \leftarrow R_2 - R_1$
$R_3 \leftarrow R_3 - R_1$
$R_4 \leftarrow R_4 - R_1$

$$\det(M) = \det \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 0 & 2 \\ 0 & 3 & 1 & 4 \end{bmatrix}$$

Expand along the first column:

$$\det(M) = 1 \cdot \det \begin{bmatrix} 1 & 1 & 2 \\ 2 & 0 & 2 \\ 3 & 1 & 4 \end{bmatrix}$$

Calculate the determinant of the $3 \times 3$ matrix:

$$\det = 1 \cdot ((0)(4) - (2)(1)) - 1 \cdot ((2)(4) - (2)(3)) + 2 \cdot ((2)(1) - (0)(3))$$ $$\det = 1 \cdot (-2) - 1 \cdot (8 - 6) + 2 \cdot (2 - 0)$$ $$\det = -2 - 1 \cdot (2) + 2 \cdot (2)$$ $$\det = -2 - 2 + 4$$ $$\det = 0$$

Since $\det(M) = 0$, the vectors are linearly dependent.
The set of vectors is $\boxed{\text{linearly dependent}}$.

A set of vectors forms a basis for $\mathbb{R}^4$ if it contains exactly four vectors and these vectors are linearly independent.

For 4.2.1: $[(0,1,0,1)', (1,1,0,0)', (0,0,1,1)']$
This set contains only 3 vectors. Since a basis for $\mathbb{R}^4$ must have 4 vectors, this set cannot be a basis for $\mathbb{R}^4$.
The set in 4.2.1 is $\boxed{\text{not a basis for } \mathbb{R}^4}$.

For 4.2.2: $[(2,2,2,1)', (2,1,1,1)', (3,2,1,1)', (1,1,1,1)']$
This set contains 4 vectors. We form a matrix $M$ with these vectors as columns and calculate its determinant.

$$M = \begin{bmatrix} 2 & 2 & 3 & 1 \\ 2 & 1 & 2 & 1 \\ 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix}$$

Swap $R_1$ and $R_4$ (this changes the sign of the determinant):

$$M' = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 2 & 1 & 2 & 1 \\ 2 & 1 & 1 & 1 \\ 2 & 2 & 3 & 1 \end{bmatrix}$$

Perform row operations:
$R_2 \leftarrow R_2 - 2R_1$
$R_3 \leftarrow R_3 - 2R_1$
$R_4 \leftarrow R_4 - 2R_1$

$$M'' = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & -1 & -1 & -1 \\ 0 & 0 & 1 & -1 \end{bmatrix}$$

Expand along the first column:

$$\det(M'') = 1 \cdot \det \begin{bmatrix} -1 & 0 & -1 \\ -1 & -1 & -1 \\ 0 & 1 & -1 \end{bmatrix}$$

Calculate the determinant of the $3 \times 3$ matrix:

$$\det = -1 \cdot ((-1)(-1) - (-1)(1)) - 0 \cdot (\dots) + (-1) \cdot ((-1)(1) - (-1)(0))$$ $$\det = -1 \cdot (1 + 1) - 0 + (-1) \cdot (-1 - 0)$$ $$\det = -1 \cdot (2) + (-1) \cdot (-1)$$ $$\det = -2 + 1 = -1$$

Since $\det(M'') = -1$, then $\det(M) = -(-1) = 1$.
Since $\det(M) = 1 \neq 0$, the vectors are linearly independent. As there are 4 linearly independent vectors in $\mathbb{R}^4$, this set forms a basis.
The set in 4.2.2 is $\boxed{\text{a basis for } \mathbb{R}^4}$.

For 4.2.3: $[(2,0,1,1)', (3,1,2,2)', (2,1,1,2)', (2,1,2,1)']$
This set contains 4 vectors. We form a matrix $M$ with these vectors as columns and calculate its determinant.

$$M = \begin{bmatrix} 2 & 3 & 2 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 2 & 1 & 2 \\ 1 & 2 & 2 & 1 \end{bmatrix}$$

Swap $R_1$ and $R_3$ (this changes the sign of the determinant):

$$M' = \begin{bmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 1 \\ 2 & 3 & 2 & 2 \\ 1 & 2 & 2 & 1 \end{bmatrix}$$

Perform row operations:
$R_3 \leftarrow R_3 - 2R_1$
$R_4 \leftarrow R_4 - R_1$

$$M'' = \begin{bmatrix} 1 & 2 & 1 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & -1 & 0 & -2 \\ 0 & 0 & 1 & -1 \end{bmatrix}$$

Expand along the first column:

$$\det(M'') = 1 \cdot \det \begin{bmatrix} 1 & 1 & 1 \\ -1 & 0 & -2 \\ 0 & 1 & -1 \end{bmatrix}$$

Calculate the determinant of the $3 \times 3$ matrix:

$$\det = 1 \cdot ((0)(-1) - (-2)(1)) - 1 \cdot ((-1)(-1) - (-2)(0)) + 1 \cdot ((-1)(1) - (0)(0))$$ $$\det = 1 \cdot (0 + 2) - 1 \cdot (1 - 0) + 1 \cdot (-1 - 0)$$ $$\det = 2 - 1 - 1$$