For Problem C-1: Our list starts with $\{1,2\}$ and $\{1,3\}$. Their complements are $\{3,4\}$ and $\{2,4\}$. Intersecting two of these four sets at a time we get the empty set and the singletons $\{1\}$, $\{2\}$, $\{3\}$, and $\{4\}$. Taking unions of two or more singletons gives all other subsets of $S$, including $S$ itself. ($S$ can also be obtained as the union of $\{1,2\}$ and its complement.)

For Problem C-2: If the unknown weights are $r,s,t,u$, we get equations $12 r + 24 s + 25 t + 25 u = 23$, etc., from the first four rows. The first four rows are the augmented matrix of this system. Row-reducing over GF(2) = $\mathbb{Z}_2$ gives

$\left[\begin{array}{rrrrr}1&0&0&0&\frac 3{20}\ 0&1&0&0&\frac 1{20}\ 0&0&1&0&\frac 7{25}\ 0&0&0&1&\frac {13}{25}\end{array}\right]$ so the weights are $r=15\%, s = 5\%, t= 28\%, u = 52\%$. Using these weights in the fifth row gives the weighted average 93.

For Problem C-3: Suppose that instead $\sqrt p \in \mathbb{Q}$, say $\sqrt p = \frac ab$ with $a,b \in \mathbb{Z}$, in lowest terms (meaning that $a$ and $b$ have no positive integer divisors in common except $1$). Then $p = \frac {a^2}{b^2}$, $a^2 = p b^2$. Since $p$ is a prime factor of $a^2$, $p$ must be a prime factor of $a$, so $a = k p$ for some integer $k$. Then $a^2 = p b^2$ becomes $k^2 p^2 = p b^2$. Cancel to get $k^2 p = b^2$. Then $p$ is a prime factor of $b^2$, and by the same reasoning as before, $p$ must be a prime factor of $b$, as it was of $a$. But then $\frac ab$ was not in lowest terms, a contradiction. We conclude that $\sqrt p \not \in \mathbb{Q}$. (We used the fact that every positive integer can be expressed uniquely as a product of primes, so $a^2$ involves the same primes as $a$ and $b^2$ involves the same primes as $b$.)

For Problem C-4:

As suggested, let $V = \mathbb{R}^2$ as a set and use the usual $+$ and $-$, but make a fudged product by scalars in which a scalar times a vector is always 0. Since every defining law of vector spaces except $1v = v$ involves a product by scalars on either both sides or neither side, all those laws are still true. However, the law $1v = v$ fails for any choice of a nonzero $v$, since the left-hand side is 0 and right-hand side is not.

For Problem C-5:

(a) The subspace spanned is the whole space, because for any $(a,b,c) \in \mathbb{R}^3$, the equation $r(1,0,0) + s(2,3,0) + t(4,5,6) = (a,b,c)$ gives $r + 2s + 4t=a$, $3s + 5t = b$, $6t = c$, which can always be solved by finding $t$, then $s$, then $r$.

(b) The subspace spanned is again the whole space, because for any $a + bx + cx^2 \in$   Pols$(\mathbb{R},2)$, the equation $r + s(2 + 3x) + t(4 + 5x + 6x^2) = a + bx + cx^2$ gives the same equations as in (a) and so can be solved.

(c) The subspace spanned is again the whole space, because for any $ax^2 + bx + c \in$   Pols$(\mathbb{R},2)$, the equation $rx^2 + s(2x + 3) + t(4x^2 + 5x + 6) = ax^2 + bx + c$ gives the same equations as in (a) and so can be solved. [If you write polynomials in some other order you get different equations but they are solved similarly.]