Let $V = \mathbb{R}_{\le 3} [x]$ with basis $ B = (1, x, x^2, x^3)$.
And $f: V \to \mathbb{R}, p \to \int_{-1}^1 p(x) dx$ and
$g: V \to \mathbb{R}^3, p \to ^t( p(-1), p(0), p(1) )$.
(1) I had to determine $M_{B \eta} (f)$ and $M_{B \eta (g)}$. [I suppose that $\eta$ means standardbasis of $\mathbb{R}$ and $\mathbb{R}^3$ in the other case! Correct me, if I am wrong. The exercise doesn't say any more]
I determined
$M_{B \eta} (f) = (2, 0, \frac{2}{3}, 0)$
$M_{B \eta} (g) = \begin{pmatrix} 1 & -1 & 1 & -1 \\1 & 0 & 0 & 0 \\1 & 1 & 1 & 1 \end{pmatrix}$.
In order to do so, I took a look at $f(1) = ... = 2, f(x) = 0, f(x^2)= 2/3, f(x^3)=0$ and $g(1) = (1,1,1), g(x) = (-1, 0, 1), g(x^2) = (1, 0, 1), g(x^3) = (-1, 0, 1)$.
Now by taking a look at those two matrices:
$M_{B \eta} (g) = ... = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \rightarrow ker(g) = span\{x - x^3\}$.
Since $f(x-x^3) = f(x) - f(x^3) = 0 - 0$ (as I computed above) it is $ker(g) \subseteq ker(f)$.
But why is $ker(g) \subset ker(f)$ ?
$ker(g)$ has dimension 2, right?
What's the span of $ker(f)$? Since $M_{B \eta} (f) = (2,0, \frac{2}{3}, 0)$ it has dimension 2, right?
We have: $p\in\ker(g)\iff p(-1)=p(0)=p(1)=0\iff p(x)=\lambda x(x+1)(x-1)=\lambda x(x^2-1)$ and we have $f(p)=0$ since $p$ is odd hence $p\in \ker (f)$. Conclude.
Remark: Notice that $f$ is a linear form so $\ker(f)$ is an hyperplan of $V$ so $$\dim\ker(f)=\dim V-1=3$$