For both parts you can simply work from the definitions.
The polynomial space in question is $P_3$, the space of polynomials of degree at most $3$. You show that $B$ spans $P_3$ by showing how to write an arbitrary $p(x)\in P_3$ as a linear combination of members of $B$, so let $p(x)=a_0+a_1x+a_2x^2+a_3x^3\in P_3$. You want to find coefficients $c_0,c_1,c_2,c_3$ such that
$p(x)=c_0\cdot1+c_1(x+1)+c_2x(x+1)+c_3x(x+1)(x-1)\;.$
Multiply out the righthand side: you want
$p(x)=(c_0+c_1)+(c_1+c_2-c_3)x+c_2x^2+c_3x^3\;.$
Now equate coefficients:
$\left\{\begin{align*} &c_0+c_1=a_0\\ &c_1+c_2-c_3=a_1\\ &c_2=a_2\\ &c_3=a_3 \end{align*}\right.\tag{1}$
Finally, show that the resulting system can always be solved for $c_0,c_1,c_2$ and $c_3$ in terms of $a_0,a_1,a_2$, and $a_3$; that assures you that every $p(x)\in P_3$ is a linear combination of elements of $B$.
Similarly, to show that $B$ is a linearly independent set, try to do what the definition of linear independence says you must do to show that a set is linearly independent: assume that $c_0,c_1,c_2$, and $c_3$ are scalars such that
$c_0\cdot1+c_1(x+1)+c_2x(x+1)+c_3x(x+1)(x-1)=0\;,$
and show that this implies that $c_0=c_1=c_2=c_3=0$. Most of the work has already been done in showing that $B$ spans $P_3$: you’re really just showing that when $a_0=a_1=a_2=a_3=0$, the system $(1)$ has the unique solution $c_0=c_1=c_2=c_3=0$, and if you’ve already done the first part, you already know that $(1)$ has a unique solution for every choice of $a_0,a_1,a_2$, and $a_3$.