Lemma: Let $X_0,\ldots,X_n$ be arbitrary random variables. Then $$\sigma(X_0,X_1-X_0,\ldots,X_n-X_{n-1}) = \sigma(X_0,X_1,\ldots,X_n).$$
Proof: The mapping $X_j-X_{j-1}$ is clearly measurable with respect to $\sigma(X_0,\ldots,X_n)$ (as difference of two measurable random variables) for all $j$. Consequently, $\sigma(X_0,X_1-X_0,\ldots,X_n-X_{n-1} \subseteq \sigma(X_0,X_1,\ldots,X_n)$. On the other hand, we can write $$X_i = \sum_{j=0}^{i-1} (X_{j+1}-X_j)+X_0;$$ this shows that $X_i$ is $\sigma(X_0,X_1-X_0,\ldots,X_n-X_{n-1})$-measurabble. This proves "$\supseteq$".
Applying this to $X_0= B(t_0)-B_0,\ldots,X_n = B(t_n)-B_0$ gives
$$\mathcal{A}_2 = \bigcup_n \sigma(B_{t_0}-B_0,\ldots,B_{t_n}-B_0)$$
where the union is taken over all $0 \leq t_0 < \ldots < t_n$ and $n \in \mathbb{N}$. Since this is $\cap$-stable generator of $\sigma(B_t-B_0; t \geq 0)$, the claim follows.
The generator of a sum of (time- and space-) homogeneous independent Markov processes is a sum of generators.
Indeed, let $X$ and $Y$ be these processes. Denote by $T_t$ and $S_t$ their Markov semigroups, $A$ and $B$ their generators, $\mathsf E_{x,y}$ the expectation given $X_0=x, Y_0=y$. Then for any $f\in C_b(\mathbb R)$ (bounded continuous), using the independence and homogeneity,
$$
\mathsf E_{x,y}[f(X_t + Y_t)] = \mathsf E_{x,y}[\mathsf E_{x,y}[f(X_t + z)]|_{z=Y_t}]= \mathsf E_{x,y}[\mathsf E_{x+z,y}[f(X_t)]|_{z=Y_t}]
\\ = \mathsf E_y[T_t f(x+z)|_{z=Y_t}] = S_t T_tf (x+y).
$$
Similarly, it equals to $T_t S_t f (x+y)$, in particular, $T_t$ and $S_t$ commute (and so do $A$ and $B$). Now for $f\in C_b^2(\mathbb R)$ (bounded twice continuously differentiable with bounded derivatives), differentiating this equality in $t$ for $t=0$, we get
$$
\frac{d}{dt}\mathsf E_{x,y}[f(X_t + Y_t)]|_{t=0} = (A+B)f(x+y),
$$
which means that $A+B$ is the generator of $X+Y$.
There is another way to look at this. A time- and space-homogeneous process is a Lévy process, which can be written as a sum
$$
at + b W_t + Z_t,
$$
where $W_t$ is a standard Wiener process, and $Z_t$ is a pure jump process. When you add two independent things of such kind, you arrive to
$$
(a_1+a_2)t + \sqrt{b_1^2 + b_2^2}\, W_t + (Z_t^1 + Z_t^2).
$$
The process $Z_t = Z_t^1 + Z_t^2$ is again a pure jump process, and the jump intensities at a given level just add up. If you translate these things to generator, everything will add up to; for instance, the diffusion part is
$\left(\sqrt{b_1^2 + b_2^2}\right)^2 \frac{d^2}{dx^2} = (b_1^2 + b_2^2)\frac{d^2}{dx^2}$.
Best Answer
Hint: You could apply Wald's identities:
In this case you can define $\mathcal{F}_t$ as $\mathcal{F}_t := \sigma\{(B_s, s \leq t),T_1\}$ (which is an admissible filtration since $T_1$ is independent of the BM) and use $\mathbb{E}T_1 = \frac{1}{\lambda}$.