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If $S$ spans $M_{22}$, then for every $2 \times 2$ matrix
$B = \begin{bmatrix}x & y \\ z & w\end{bmatrix}$,
there exist constants $\alpha, \beta, \gamma$ so that
\begin{align*}
\begin{bmatrix}x & y \\ z & w\end{bmatrix}
&=
\alpha\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix} +
\beta\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix} +
\gamma\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}
\end{align*}
Applying Definition ME, this leads to the linear system
\begin{align*}
\alpha + 2\beta &= x\\
2\alpha + \beta + \gamma &= y\\
2\alpha - \beta + \gamma &= z\\
\alpha + 2\beta + 2\gamma &= w.
\end{align*}
We need to row-reduce the augmented matrix of this system by hand due to the symbols $x$, $y$, $z$, and $w$ in the vector of constants.
\begin{align*}
\begin{bmatrix}
1 & 2 & 0 & x\\
2 & 1 & 1 & y\\
2 & -1 & 1 & z\\
1 & 2 & 2 & w
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 0 & x - y + z\\
0 & \leading{1} & 0 & \frac{1}{2}(y - z)\\
0 & 0 & \leading{1} & \frac{1}{2}(w - x)\\
0 & 0 & 0 & \frac{1}{2}(5y - 3x - 3z - w)
\end{bmatrix}
\end{align*}
It is easy to find values of $x, y, z$ and $w$ so that the final column will become a pivot column after a few more row operations, namely any choice with $5y - 3x - 3z - w \ne 0$. Thus, by Theorem RCLS the system would be inconsistent and there is a matrix in $M_{22}$ that is not in the span of $S$. One such matrix is $B = \begin{bmatrix} 3 & 3 \\ 3 & 2\end{bmatrix}$.) So $S$ does not span $M_{22}$.
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To check for linear independence, we need to see if there are nontrivial coefficients $\alpha, \beta, \gamma$ that solve
\begin{align*}
\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
&=
\alpha\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix} +
\beta\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix}+
\gamma\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}
\end{align*}
This requires the same work that was done in part (a), with $x = y = z = w = 0$.
In that case, the coefficient matrix row-reduces so the first three columns are pivot columns and a row of zeros on the bottom, so we know that the only solution to the matrix equation is $\alpha = \beta = \gamma = 0$. So the set $S$ is linearly independent.