1. Find set $A$

We need to solve

But note that $x+4=(x+3)+1.$ For any real $t$, $[t+1]=[t]+1.$ So, $[x+4]=[x+3]+1.$ Hence $[x+3]+[x+4]=[x+3]+([x+3]+1)=2[x+3]+1.$

Therefore the inequality becomes $2[x+3]+1\u00d63$ 2[x+3]6 [x+3]1.

Since [x+3]1 implies $x+3<2$ $x<-1.$ Also, if $x<-1$, then indeed $x+3<2$, so [x+3]1.

Thus,

2. Find set $B$

Given $B=\left\{x\in\mathbb R:3^x\left(\sum_{r=1}^{\infty}\frac{3}{10^r}\right)^{x-3}<3^{-3x}\right\}.$

First evaluate the series: $\sum_{r=1}^{\infty}\frac{3}{10^r}=3\sum_{r=1}^{\infty}\left(\frac{1}{10}\right)^r.$ This is a geometric series with first term $\frac{1}{10}$ and common ratio $\frac{1}{10}$, so $\sum_{r=1}^{\infty}\left(\frac{1}{10}\right)^r=\frac{\frac{1}{10}}{1-\frac{1}{10}}=\frac{1}{9}.$ Hence $\sum_{r=1}^{\infty}\frac{3}{10^r}=3\cdot\frac{1}{9}=\frac{1}{3}.$

So the inequality becomes $3^x\left(\frac13\right)^{x-3}<3^{-3x}.$ Now $\left(\frac13\right)^{x-3}=3^{-(x-3)}=3^{-x+3}.$ Thus left side is $3^x\cdot 3^{-x+3}=3^3=27.$ Therefore, $27<3^{-3x}.$ Since $27=3^3$, we get $3^3<3^{-3x}.$ Because base $3>1$, compare exponents: $3<-3x$ $x<-1.$

Thus,

3. Compare $A$ and $B$

We found A=(-8),\qquad B=(-8). Therefore, $A=B.$

So the correct option is: $\boxed{\text{C}}$

4. Compare with stored correct answer

Stored correct answer: $\text{C}$.

Our derived answer is also $\text{C}$, so they agree.