For the first, let $\phi(a,b,c) = \int_{-1}^{1} | x^{3}-a-bx-cx^{2}| dx$. Notice that $\phi(a,b,c) = \phi(-a,b,-c)$ and since $\phi$ is convex, we have $\phi(0,b,0) \leq \frac{1}{2} (\phi(a,b,c) +\phi(-a,b,-c))$. Hence we can assume that $a=c=0$ and the problem reduces to minimizing $\xi(b) = \int_{-1}^{1} | x^{3}-bx| dx = 2\int_{0}^{1} x| x^{2}-b| dx$. If $b<0$, we have $| x^{2}-b| = x^2-b>x^2$, hence $\xi(b) \geq \xi(0)$. So we can assume that $b \geq 0$. Similarly, if $b>1$, we have $| x^{2}-b| =b-x^2 > 1-x^2 = |x^2-1|$ over the range of integration. Hence $\xi(b) \geq \xi(1)$. Hence we may assume that $b \in [0,1]$. Evaluating the integral gives $\xi(b) = 2\frac{1}{4}(2 b^2-2b+1) $, and minimizing with respect to $b$ over $[0,1]$ gives $b = \frac{1}{2}$, and evaluating gives $\xi(\frac{1}{2}) = \frac{1}{4}$.
For the second, letting $p_k(x) = x^k$, and choosing the space as $L_2[-1,1]$ lets us write the problem as $\sup \{ \langle p_3, g \rangle | \langle p_k, g \rangle = 0 , k=0,1,2, \ \ \|g\| = 1\}$. If we let $S = \text{sp} \{p_k\}_{k=0}^2$ (which is closed, since it is finite dimensional), the problem can be written as $\sup_{g \in S^\bot, \|g\| = 1} \langle p_3, g \rangle$. Since the function $g \mapsto\langle p_3, g \rangle$ is linear,this problem is the same as $\sup_{g \in S^\bot, \|g\| \leq 1} \langle p_3, g \rangle$. Now let $\Pi$ be the orthogonal projection onto $S^\bot$, then we can write the problem as $\sup_{\|\Pi g\| \leq 1} \langle p_3, \Pi g \rangle$, and since $\|\Pi g\| \leq \|g\|$, we see that this is equal to $\sup_{\| g\| \leq 1} \langle p_3, \Pi g \rangle = \sup_{\| g\| \leq 1} \langle \Pi^* p_3, g \rangle = \|\Pi p_3\|$, since $\Pi$ is self-adjoint. Let $P_n$ be the Legendre polynomials, then we see that $p_3 = \sum_{k=0}^3 \alpha_k P_k$, and $S = \text{sp} \{P_k\}_{k=0}^2$. Since the $P_n$ are orthogonal, we see that $\Pi p_3 = \alpha_3 P_3$, and the solution to the problem is $|\alpha_3\|\|P_3\|$. Since $P_3(x) = \frac{1}{2}(5 x^3-3x)$, we see that we must have $\alpha_3 = \frac{2}{5}$. Since $\|P_3\| = \sqrt{\frac{2}{7}}$, we see that the answer is $|\alpha_3\|\|P_3\| = \frac{2}{5} \sqrt{\frac{2}{7}} = \sqrt{\frac{8}{175}}$.
(Addendum: Note that a minimizing $g$ is $g = \frac{1}{\|\Pi p_3\|} \Pi p_3 = \frac{1}{\|P_3\|} P_3 = \sqrt{\frac{7}{2}}\frac{1}{2}(5 p_3-3p_1)$, or $g(x) = \sqrt{\frac{7}{8}}(5 x^3-3x)$.)