A partial result: any FPA mus be congruent to $2, 3 \bmod 5$, and these primes have density $\frac{1}{2}$ in the set of all primes by the strong form of Dirichlet's theorem.
First, I claim that Binet's formula $F_n = \frac{\phi^n - \varphi^n}{\phi - \varphi}$ (my numbering differs from yours), where $\phi, \varphi$ are the two roots of $x^2 = x + 1$, is valid modulo $p$ for any prime $p \neq 5$, where $\phi, \varphi$ are interpreted as elements of the splitting field of $x^2 = x + 1$ over $\mathbb{F}_p$. The proof is the same proof as in the usual case, the point being that the discriminant of $x^2 = x + 1$ is $5$, so for any other prime there are two distinct roots. Moreover, since $(\phi - \varphi)^2 = 5$, we know that $\phi - \varphi \neq 0$ for any such prime. It follows that $F_n \equiv 0 \bmod p$ if and only if $\phi^n \equiv \varphi^n \bmod p$.
If $\left( \frac{5}{p} \right) = 1$, then $x^2 = x + 1$ splits over $\mathbb{F}_p$, from which it follows that $\phi^{p-1} \equiv \varphi^{p-1} \equiv 1 \bmod p$, hence that
$$F_{p-1} \equiv 0 \bmod p.$$
By quadratic reciprocity, $\left( \frac{5}{p} \right) = \left( \frac{p}{5} \right)$, hence these are precisely the primes $p \equiv 1, 4 \bmod 5$. So none of these primes are FPAs.
If $\left( \frac{5}{p} \right) = -1$, then $x^2 = x + 1$ has splitting field $\mathbb{F}_{p^2}$. The Galois group is generated by the Frobenius map $x \mapsto x^p$, hence $\phi^p \equiv \varphi \bmod p$ and, similarly, $\varphi^p \equiv \phi \bmod p$, hence $\phi^{p+1} \equiv \phi \varphi \equiv \varphi^{p+1} \equiv -1$, from which it follows that
$$F_{p+1} \equiv 0 \bmod p.$$
This occurs if and only if $p \equiv 2, 3 \bmod 5$.
Edit: Second partial result: any FPA must be congruent to $3 \bmod 4$. Now the density has been reduced to $\frac{1}{4}$. To see this, note that since $\phi \varphi = -1$, the condition that $\phi^n \equiv \varphi^n \bmod p$ is equivalent to the condition that $(-\phi^2)^n \equiv 1 \bmod p$. On the other hand, we know that when $p \equiv 2, 3 \bmod 5$ we have $\phi^{p+1} \equiv -1 \bmod p$, hence
$$\left( - \phi^2 \right)^{ \frac{p+1}{2} } \equiv (-1)^{ \frac{p-1}{2} } \bmod p.$$
It follows that when $p \equiv 1 \bmod 4$ we have
$$F_{ \frac{p+1}{2} } \equiv 0 \bmod p.$$
I don't expect these necessary conditions to be sufficient. The question is similar to asking that $- \phi^2$ behave like a primitive root, so this question or a variation on it may end up being an open problem.
Edit #2: For example, you can verify for yourself that $F_{16} \equiv 0 \bmod 47$ even though $47 \equiv 2 \bmod 5$ and $47 \equiv 3 \bmod 4$.
Each even number $n$ is of the form $n=2^ku$ where $u$ is odd and $k\ge 1$. If $k=1$ then $n$ is already an E-prime. Otherwise let the first E-prime be $2u$ and all others be $2$ as many as needed.
Note this method relies on a small part of unique factorization, and it is not the same as the way suggested to solve it in the Silverman text.
Note: For part A you may want to be more explicit and say an E-prime is any number of the form $2u$ with $u$ odd. Also it may or may not need a sign, depending on whether the E-zone includes negative integers.
Elaboration of part B argument: First, the statement that
Every even number $n$ is of the form $2^ku$ with $u$ odd and $k\ge 1$
can be shown by induction. Base case $n=2=2\cdot 1$ is of the desired form with $k=1,u=1$. Now suppose $n>2$ is even so that $n=2k$ for some $k>1$. If $k$ is odd then we have the desired form $n=2^1u$ where $u=k$. If $k$ is even, then $k=2k'$ and then $n=2\cdot(2k').$ At this point $2k'$ is an even number less than $n$ so by the inductive hypotheses $2k'=2^tu$ with $t\ge 1$ and $u$ odd. Putting this back into $n=2 \cdot 2k'$ gives $n=2^{t+1}u$ which is of the desired form, finishing the inductive step.
Now that's done, and you want to show every even number $n$ is a product of E-primes, where these are all numbers of the form $2u$ with $u$ odd (from part A). Note first that $2=2\cdot 1$ qualifies as an E-prime using $u=1$. Since $n$ is even we can use the above shown fact and write $n=2^ku$ with $k\ge 1$ and $u$ odd. There are two cases:
case 1: $k=1$. In this case $n=2^1u=2u$ with $u$ odd, so $n$ is itself an E-prime. It is thus the product of a single E-prime and the result holds in this case.
case 2: $k>1$. Here we can write $k=1+r$ and then
$$n=2^ku=(2u)\cdot 2 \cdot 2 \cdots \cdot 2,$$
where there are $r$ copies of the E-prime $2$ being multiplied together after the initial E-prime $(2u)$. So in case 2 also, we have $n$ as a product of E-primes.
I hope this helps explain part B...
Best Answer
An even integer $2n$ is prime in the $\Bbb{E}$-zone if and only if $2n \equiv 2 \pmod{4}$, which occurs if and only if $n$ is odd. So, factorization into primes amounts to writing $$ \begin{align} N &= 2n_1 \cdot 2n_2 \cdots 2n_k \\ &= 2^k \cdot n_1 \cdot n_2 \cdots n_k, \end{align} $$ where $n_1, \ldots, n_k$ are odd integers. Since we're looking for small numbers with many prime factorizations, and since the product of odd numbers is odd, we may as well assume that $k = 2$. (You seem to have done this implicitly by writing prime factorizations as pairs.)
If we want $4$ prime factorizations, then we're looking for $4$ pairs of prime factors, or an odd number with $8$ factors. (The factors are automatically odd, too.)
In the integers, the number of factors of the number $p_1^{e_1} \cdots p_r^{e_r}$, where $p_1, \ldots, p_r$ are usual primes, is $$ (1 + e_1) \cdots (1 + e_r). $$ So, we need to write $8$ (the number of factors we desire) as $2 \cdot 2 \cdot 2$ or as $4 \cdot 2$. We will consider each possibility in turn.
In the first case, we must consider a product of distinct (odd!) primes $p_1 \cdot p_2 \cdot p_3$. The smallest example of this is $3 \cdot 5 \cdot 7 = 105$, whose factors can be paired off as: $$ \begin{array}{ccc} 1 && 3 \cdot 5 \cdot 7 \\ 3 && 5 \cdot 7 \\ 5 && 3 \cdot 7 \\ 7 && 3 \cdot 5 \end{array} $$ Back in the $\Bbb{E}$-zone, these are prime factorizations of $$ 2^2 \cdot 3 \cdot 5 \cdot 7 = 420 $$ (by putting a $2$ on each factor of the pair): $$ \begin{array}{ccc} 2 && 210 \\ 6 && 70 \\ 10 && 42 \\ 14 && 30 \end{array} $$
In this case, consider a number of the form $p_1^3 \cdot p_2$. The smallest example of this is $3^3 \cdot 5 = 135$, which is larger than $105$, but I'll follow through the analysis anyway. The factors in pairs are: $$ \begin{array}{ccc} 1 && 3^3 \cdot 5\\ 3 && 3^2 \cdot 5 \\ 3^2 && 3 \cdot 5 \\ 3^3 && 5 \end{array} $$ Back in the $\Bbb{E}$-zone, these are prime factorizations of $$ 2^2 \cdot 3^3 \cdot 5= 540 $$ (by putting a $2$ on each factor of the pair): $$ \begin{array}{ccc} 2 && 270 \\ 6 && 90 \\ 18 && 30 \\ 54 && 10 \end{array} $$
So, $420$ is the smallest positive integer with $4$ prime factorizations in the $\Bbb{E}$-zone.