$ S_3 = \{w1, w2, w3, w4\}, w1 = \left(\begin{array}\ 2 \\ 1 \\ 0 \\ 2 \\ \end{array}\right), w2 = \left(\begin{array}\ 1 \\ 0 \\ 2 \\ 2 \\ \end{array}\right), w3 = \left(\begin{array}\ 0 \\ 2 \\ 2 \\ 1 \\ \end{array}\right), w4 = \left(\begin{array}\ 2 \\ 2 \\ 1 \\ 0 \\ \end{array}\right)$
$ S_1 = \{u1, u2, u3\}, u1 = \left(\begin{array}\ 1 \\ 0 \\ 2 \\ 1 \\ \end{array}\right), u2 = \left(\begin{array}\ 0 \\ 2 \\ 1 \\ 2 \\ \end{array}\right), u3 = \left(\begin{array}\ 1 \\ 1 \\ 0 \\ 2 \\ \end{array}\right) $
(1) Is $ span(S_3)$ = $ R^4 $ ?
$ \quad $ Yes, the linear system has no zero row, is an identity matrix, therefore it is always consistent.
(2) $ Which\ of\ the\ vectors\ in\ S_3\ belongs\ to\ span(S_1)\ $ ?
$ \quad \quad $ None (all linear systems are found to be inconsistent)
(3) What is the intersection of $ span(S_3)\ and\ span(S_1) $ ?
$ \quad \quad span(S_3) \cap span(S_1)= R^4 \cap span(S_1) = span(S_1) $
I am not sure if my answers are correct for (2) and (3), if $S_3$ spans the entire $R^4$ space, would there not be at least some vectors that can be expressed as a linear combination of the vectors in $ S_1 $ for part (2), but the linear systems suggest otherwise? Please correct me if I am wrong. Thank you.
(3) is correct.
The answer to (1) is correct (the span is all of $\Bbb R^4$), but I don't understand your justification—what is the identity matrix you refer to?
As for (2), I will trust your computations that none of the vectors in $S_3$ are in the span of $S_1$. As for the apparent paradox you noticed: your answer to (3) shows that lots of vectors in the span of $S_3$ are in the span of $S_1$, but that's not the same as saying that vectors from $S_3$ itself must be in the span of $S_1$.
For a simpler example, in $\Bbb R^2$: every vector in the span of $\{(1,0)\}$ is also in the span of $\{ (1,1) ,(1,-1) \}$, even though neither $(1,1)$ nor $(1,-1)$ is in the span of $\{(1,0)\}$.