Happy new year!

There are a total of $2^{11} = \boxed{2048}$ strings of length 11. We now have to subtract the strings that don't meet the conditions. In the following, I will refer to the repeated letters as the row .

There are 4 different places where exactly 8 consecutive same letters can occur; from position 1–8, 2–8, 3–10 or 4–11, and 2 choices for the letter in the row. For 1–8 and 4–11, the letter next to the row has to be different and for the two remaining letters, there are $2^2 = 4$ choices, so for 1–8 and 4–11 there are $2 \cdot 2^2 + 2 \cdot 2^2 = \boxed{16}$ strings. For 2–9 and 3–10, the letter before and after the row have are fixed (so that the row ends there), but the remaining letter has 2 options, so there are a total of $2 + 2 = \boxed{4}$ sequences with the row at 2–9 or 3–10.

We now do the same for a row of exactly 9 $A$ 's. The row can be at 1–9, 2–10 or 3–11 and there are 2, 1 and 2 such sequences respectively, so a total of $2+1+2=\boxed{5}$ .

Therefore, and since both conditions can't happen at the same time, the overall number of strings that meet both conditions is

$2048 - 16 - 4 - 5 = \boxed{\boxed{\huge 2019}}$

$\huge\mathbb{H}\text{appy new }\mathbb{Y}\text{ear!!}$