In this hexahedron, define the planes $\pi_0 = ABCD$, $\pi_1 = ABFE$, $\pi_2 = BCGF$ , $\pi_3 = CDHG$ , $\pi_4 = DAEH$, $\pi_5 = EFGH $. You are given that $ABCD$ is an isosceles trapezoid, with $AB = 10, CD = 4 $, and $AD = BC = 3\sqrt{5} $. In addition, $\angle(\pi_0, \pi_1) = 70^\circ, \angle(\pi_0, \pi_2)= \angle(\pi_0, \pi_4) = 80^\circ, \angle(\pi_0, \pi_3) = 90^\circ$, and finally, you're given that $AE = BF = CG= DH = 5$. Determine the side lengths $EF, FG, GH, HE$ and $\angle FBA$.
My thoughts:
The equations of all the planes are possible to find, and then their intersections can be found numerically. Then the coordinates of $E,F,G,H$ can be evaluated and hence the side lengths can be computed. Similarly $\angle FBA$ can be found from the vectors $\vec{BF}$ and $\vec{BA}$.
I'm going to use your idea.
We may suppose that the equation of $\pi_0$ is $z=0$ with $A(-5,0,0),B(5,0,0),C(2,6,0),D(-2,6,0)$ and $E_z,F_z,G_z,H_z\gt 0$.
Then, we have
$$\begin{align}&\pi_1 : y-\cot(70^\circ)z=0 \\\\&\pi_2 : 2x+y+\sqrt 5\cot(80^\circ)z-10=0 \\\\&\pi_3 : y=6 \\\\&\pi_4 : 2x-y-\sqrt 5\cot(80^\circ)z+10=0\end{align}$$
Letting $$E(a,b,c),F(d,e,f),G(g,h,i),H(j,k,l)$$ we have the following system : $$\begin{cases}2a-b-\sqrt 5pc+10=0 \\\\b-qc=0 \\\\2d+e+\sqrt 5pf-10=0 \\\\e-qf=0 \\\\h=6 \\\\2g+h+\sqrt 5pi-10=0 \\\\k=6 \\\\2j-k-\sqrt 5pl+10=0 \\\\\sqrt{(a+5)^2+b^2+c^2}=5 \\\\\sqrt{(d-5)^2+e^2+f^2}=5 \\\\\sqrt{(g-2)^2+(h-6)^2+i^2}=5 \\\\\sqrt{(j+2)^2+(k-6)^2+l^2}=5 \\\\c,f,i,l\gt 0\end{cases}$$ where $p:=\cot(80^\circ)$ and $q:=\cot(70^\circ)$.
Solving the system gives $$\begin{align}a&=-5+\frac{5(q+\sqrt 5 p)}{\sqrt{5p^2+2\sqrt 5pq+5q^2+4}} \\\\b&=\frac{10q}{\sqrt{5p^2+2\sqrt 5pq+5q^2+4}} \\\\c&=\frac{10}{\sqrt{5p^2+2\sqrt 5pq+5q^2+4}} \\\\d&=5-\frac{5(q+\sqrt 5 p)}{\sqrt{5p^2+2\sqrt 5pq+5q^2+4}} \\\\e&=\frac{10q}{\sqrt{5p^2+2\sqrt 5pq+5q^2+4}} \\\\f&=\frac{10}{\sqrt{5p^2+2\sqrt 5pq+5q^2+4}} \\\\g&=2-\frac{5\sqrt 5p}{\sqrt{5p^2+4}} \\\\h&=6 \\\\i&=\frac{10}{\sqrt{5p^2+4}} \\\\j&=-2+\frac{5\sqrt 5p}{\sqrt{5p^2+4}} \\\\k&=6 \\\\l&=\frac{10}{\sqrt{5p^2+4}}\end{align}$$ where $b=e,c=f,i=l,h=k,g+j=0$ and $a+d=0$ hold.
Therefore, we finally have $$EF=|a-d|=|-d-d|=2d$$
$$GH=|g-j|=|g-(-g)|=2g$$
$$FG=\sqrt{(d-g)^2+(e-h)^2+(f-i)^2}=\sqrt{(j-a)^2+(k-b)^2+(l-c)^2}=HE$$
$$\angle{FBA}=\arccos\bigg(\frac{5-d}{\sqrt{(d-5)^2+e^2+f^2}}\bigg)$$