The wonderful booklet Borel Spaces by Rao and Rao gives a number of examples. Here is one of them:
Lemma: There is an injective function $f:[0,1]\to [0,1]$ such that the preimage of
every uncountable Borel set with uncountable complement is not a Borel set.
Proof: There are continuum many Borel sets, so let $(A_0,B_0), (A_1,B_1),\ldots, (A_\alpha,B_\alpha),\ldots,$ with $\alpha<\mathfrak{c}$ be a well-ordering of all disjoint pairs of uncountable Borel sets. Constuct recursivelya $\mathfrak{c}$-sequence of pairs by choosing for each $\beta<\mathfrak{c}$ two points $x_\beta\in A_\beta$ and $y_\beta\in B_\beta$ that are not in $\{x_\alpha, y_\alpha;\alpha<\beta\}$ (this is possible since every uncountable Borel set has the cardinality of the continuum). Let $f(x_\alpha)=y_\alpha$, $f(y_\alpha)=x_\alpha$ and let $f(z)=z$ for all $z$ not occuring in the sequence. We now verify that $f$ has the required property. It is clear that $f$ is injective. Let $A$ be an uncountable Borel set with uncountable complement. Since there are uncountably many pairs of disjoint uncountable Borel subsets of $A$, we have that $D=f^{-1}(A)\cap A$ is uncountable. It has an uncountable complement, because $A$ has one. Suppose that $f^{-1}(A)$ is a Borel set. Then $D$ is an uncountable Borel set with uncountable complement, such that $D=f^{-1}(D)$. But this is impossible. The pair $(D,D^C)$ must show up in our well-ordering, so $f$ maps a point in $D$ to a point in $D^C$. $\blacksquare$
Corollary: Let $f$ be an injective function $f:[0,1]\to [0,1]$ such that the preimage of every uncountable Borel set with uncountable complement is not a Borel set. Let $\mathcal{D}$ be the $\sigma$-algebra generated by $f$ and $\mathcal{B}$ be the Borel $\sigma$-algebra. Then $\mathcal{D}\cap\mathcal{B}$ is the $\sigma$-algebra consisting of countable sets and sets with countable complements.
Proof: Every, set in $\mathcal{D}$ that is uncountable and has an uncountable complement must be the preimage of an uncountable Borel set with uncountable complement. But such preimages are by the last lemma never Borel sets. So every set in $\mathcal{D}\cap\mathcal{B}$ is countable or has a countable complement. Moreover, it is easy to see that every such set is in the intersection. $\blacksquare$
We show that the $\sigma$-algebra on $[0,1]$ consisting of countable sets and sets with countable complements is not countably generated (Rao & Rao prove this in a different way). First, we need a Lemma:
Lemma: Let $\mathcal{C}$ be a countably generated $\sigma$-algebra on a set $X$. Then there is a function $g:X\to [0,1]$ such that $\mathcal{C}$
is the $\sigma$-algebra generated by $g$ (where the codomain is endowed with the Borel $\sigma$-algebra). Moreover, if for all distinct $x,y\in X$ there is a $C\in\mathcal{C}$ such that $x\in C$ and $y\notin C$, the $g$
must be injective.
Proof: Let $\{C_0,C_1,C_2,\ldots\}$ be a countable set of generators. Let $g$ be given by $$g(x)=\sum_{n=0}^\infty\frac{2}{3^{n+1}}I_{C_n},$$ where $I_{C_n}$
is the indicator function of $C_n$. The function $g$ is known as the Marczewski-function. If $g(x)=g(y)$, then there is no nonempty set $B$ such that
$x\in g^{-1}(B)$ but $y\notin g^{-1}(B)$. $\blacksquare$
Corollary: The $\sigma$-algebra on $[0,1]$ consisting of countable sets and those with countable complement is not countably generated.
Proof: Suppose it is. Then by the preceding lemma, the $\sigma$-algebra is generated by some $g:[0,1]\to[0,1]$. Let $x,y\in[0,1]$ be two different points. Then $x\in [0,1]\backslash\{y\}$ but $y\notin[0,1]\backslash\{y\}$. It follows that $g$ must be injective. Since $g$ is measurable, either $g^{-1}\big([0,1/2]\big)$ or $g^{-1}\big([1/2,1]\big)$ must have a countable complement. Say, it is is the latter. Again $g^{-1}\big([1/2,3/4]\big)$ or $g^{-1}\big([3/4,1]\big)$ must have countable complement. Continuing this way, we get a decreasing sequence of sets with countable complement. It follows that the intersection $I$ has a countable complement. But $I$ is the preimage of a singleton under $g$. This contradiction proves that the $\sigma$-algebra is not countably generated. $\blacksquare$
First define
$$\mathscr{D}_1 = \{A_1 \in \sigma(\mathscr{A}_1): \forall A_2 \in \mathscr{A}_2, P(A_1 \cap A_2) = P(A_1)P(A_2) \}.$$
By condition, $\mathscr{A}_1 \subset \mathscr{D}_1$. Also note
As an algebra, $\mathscr{A}_1$ is a $\pi$-system (thus the original condition can be weakened).
$\mathscr{D}_1$ is a $\lambda$-system, by the countable additivity property of probability measures.
Thus by Dynkin's $\pi$-$\lambda$ theorem, we have $\sigma(\mathscr{A}_1) \subset \mathscr{D}_1$ (i.e., $\sigma(\mathscr{A}_1) = \mathscr{D}_1$).
Next define
$$\mathscr{D}_2 = \{A_2 \in \sigma(\mathscr{A}_2): \forall A_1 \in \sigma(\mathscr{A}_1), P(A_1 \cap A_2) = P(A_1)P(A_2) \}.$$
By what was shown above, we know $\mathscr{A}_2 \subset \mathscr{D}_2$. Using the similar trick as in
Step 1, one can show that $\sigma(\mathscr{A}_2) =\mathscr{D}_2$, which proves your statement.
Best Answer
In the second case $\{-n\} =F_{m+1}-F_m$ where $m=2n-1$. I will you verify that $\{n\}$ is also in the sigma algebra for $n \geq 0$. Hence every singleton set is in $\sigma (\mathcal F)$ which makes $\sigma (\mathcal F)$ the power set of $\mathbb Z$.
$\sigma (\mathcal E)$ is the class of al symmetric subsets of $\mathbb Z$.