Let's look at a proof of (1) again.
The key is that $10 \equiv 1 \pmod 9$. If we write the number
$$n = a_k 10^k + a_{k-1}10^{k-1} + \ldots + a_0 = a_k a_{k-1} \ldots a_0,$$ where the last expression is digit-by-digit and not multiplication and $0 \leq a_i < 10$, then we get that
$$a_k a_{k-1} \ldots a_0 \pmod 9$$
is the same as
$$a_k 10^k + a_{k-1}10^{k-1} + \ldots + a_0 = a_k a_{k-1} \ldots a_0 \equiv a_k + a_{k-1} + \ldots + a_0 \pmod 9,$$
which is the statement we were asked to prove. But they key was that $10 \equiv 1 \pmod 9$, and that we happen to write in a base-10 system, so that the way we express each digit is with a power of $10$ in front of it. The exact same proof works for remainders upon division by $3$, since $10 \equiv 1 \pmod 3$ too.
Now suppose we don't break our number into one-digit blocks, but instead two-digit blocks. We are essentially using a base-100 system now, as silly as that sounds. Before, the key was that $10 \equiv 1 \mod m$. For what $m$ is $100 \equiv 1 \mod m$? Answer: when $m = 3, 9, 11$, i.e. the factors of $99$. (If you know the Chinese Remainder Theorem or have done enough arithmetic to get a good intuition, this may feel very obvious).
So I claim that writing
$$n = b_k 100^k + b_{k-1}100^{k-1} + \ldots + b_0,$$
where this time each $b_i$ satisfies $0 \leq b_i < 100$ (i.e. we split the number into 2-digit chunks), then $n \equiv \sum b_i \pmod{m}$, where $m = 3,9,$ or $11$. Why? Let's check to make sure we understand. We'll check with $11$. We know that $100 \equiv 1 \pmod{11}$, so that $100^k \equiv 1 \pmod{11}$. Then
$$b_k 100^k + b_{k-1}100^{k-1} + \ldots + b_0 \equiv b_k + b_{k-1} + \ldots + b_0 \pmod{11}.$$
And so we have it, and we've developed an easier-to-compute-maybe divisibility test.
The key part for the alternating sum in (2) is that $10 \equiv -1 \pmod{11}$, so that if
$$n = a_k 10^k + a_{k-1}10^{k-1} + \ldots + a_0,$$
then $n \equiv a_0 - a_1 + a_2 - \ldots + (-1)^k a_k \pmod{11}$. So for two-digit chunks in an effective base-100 system, we get the same sort of behavior for the divisors of $101$. But wouldn't you know - $101$ is prime. So all we can conclude that $n \equiv \sum (-1)^jb_j \pmod {101}$.
Similarly, for 3-digit splits, we can conclude that for the factors $m$ of $999$, $n \equiv \sum c_j \pmod m$, and for the factors $d$ of $1001$, $n \equiv \sum (-1)^j c_j \pmod d$. (I continue my pattern, $c_j$ comes from three-digit blocks).
In fact, I happen to know that $1001 = 7 \cdot 11 \cdot 13$, and the take-alternating-sum-and-difference-of-three-digit-blocks is actually how I quickly decide if a number is divisible by $7, 11, 13$ all the time.
