As saulspatz says, your approximation is already good, but if you want an exact answer:
For (b), the probability that a single person wins at least twice (because if they win 3 or more times, they have still won twice) can be calculated as the total probability minus the probability of losing every time or winning only once:
$$1-\dbinom{1000}{0}\left(\dfrac{1}{\dbinom{49}{6}}\right)^{0}\left(1-\dfrac{1}{\dbinom{49}{6}}\right)^{1000} - \dbinom{1000}{1}\left(\dfrac{1}{\dbinom{49}{6}}\right)^{1}\left(1-\dfrac{1}{\dbinom{49}{6}}\right)^{999} \approx 2.55\times 10^{-9}$$
which is basically what you found, as well.
For (c), if we let $p$ be the probability of winning at least twice found above, we can do the same thing. The probability of at least one person winning at least twice is one minus the probability of no one winning at least twice gives the probability of at least one person winning twice.
$$1-(1-p)^{60000000} \approx 0.142089$$
So, your estimation was very close.
It can be solved by applying the rule: $$\text{probability=number of favorable outcomes divided by number of possible outcomes}$$This rule works if the outcomes are equiprobable (which is the case below).
It is IMV the best way to prove this problem rigorously.
For convenience number the tickets with $1,2,\dots,N$ and let the tickets with a number $\leq M$ be the winning tickets.
Fix one of the persons that draws a ticket.
For $i=1,2,\dots,N$ let $E_i$ denote the event that ticket $i$ is drawn by this person.
If $i$ and $j$ are distinct ticket numbers then there is no reason at all to think that the ticket with number $i$ has more (or less) chance than the ticket with number $j$ to become the ticket that is drawn by this person.
In short: $$P(E_i)\text{ does not depend on }i\tag1$$
Next to that is for sure that one of the tickets is drawn by the person so that:$$\sum_{i=1}^NP(E_i)=1\tag2$$
From $(1)$ and $(2)$ we conclude for $i=1,\dots,N$ that: $$P(E_i)=\frac1N$$
Then consequently if $E$ denotes the event that the person draws a winning ticket:$$P(E)=P\left(\bigcup_{i=1}^ME_i\right)=\sum_{i=1}^MP(E_i)=\frac{M}N$$
Edit (concerning your efforts):
"Let's suppose it's person #$K$'s turn to draw and he draws some ticket $A$...."
Under that supposition $H_1$ and $B$ (as defined in your question) denote exactly the same event so that directly:$$P(B)=P(H_1)$$
The application of the total law of probability is not incorrect but is redundant (hence confusing).
"Obviously the probability of each ticket being a winning ticket is $M/N$..."
Indeed, and that tells us that: $$P(H_1)=\frac{M}N$$
So your end result is:$$P(B)=\frac{M}{N}$$ which agrees with the answer that I provided above.
IMV your approach is correct but is up to some level tarnished by the redundant use of the total probability law.
Best Answer
It would appear that the dominoes $B$ has to choose from are the $21$ without any blank faces. By my calculation this would produce the relative chances given as the answer.
Following the procedure suggested by user2661923 in the comments, let $\ f_0=1\ $, $\ f_1\ $ be the probability that no-one wins in round $1$, and, for $\ i=2,3,\dots,6\ $, $\ f_i\ $ the conditional probability that no-one wins in round $\ i\ $ given that no-one has won in any of the previous $\ i-1\ $ rounds. Then \begin{align} f_i&=\frac{5(6-i)(19-i)}{6(7-i)(22-i)}\\ P\big(\text{$A$ wins in round $i$}\big)&=\frac{\prod_\limits{k=0}^{i-1}f_k}{7-i}\ ,\\ P\big(\text{$A$ wins}\big)&=\sum_{i=1}^6\frac{\prod_\limits{k=0}^{i-1}f_k}{7-i}\ ,\\ P\big(\text{$B$ wins in round $i$}\big)&=\frac{3(6-i)\prod_\limits{k=0}^{i-1}f_k}{(22-i)(7-i)}\ ,\\ P\big(\text{$B$ wins}\big)&=\sum_{i=1}^6\frac{3(6-i)\prod_\limits{k=0}^{i-1}f_k}{(22-i)(7-i)}\ ,\\ P\big(\text{$C$ wins in round $i$}\big)&=\frac{(19-i)(6-i)\prod_\limits{k=0}^if_k}{6(22-i)(7-i)}\ ,\\ P\big(\text{$C$ wins}\big)&=\sum_{i=1}^6\frac{(19-i)(6-i)\prod_\limits{k=0}^if_k}{6(22-i)(7-i)}\ .\\ \end{align} Here is some Magma code to carry out the calculation:
If you copy and paste this into the online Magma calculator it returns the values \begin{align} P\big(\text{$A$ wins}\big)&=\frac{773347}{1551312}\\ P\big(\text{$B$ wins}\big)&=\frac{14789}{57456}= \frac{399303}{1551312}\\ P\big(\text{$C$ wins}\big)&=\frac{189331}{775656}=\frac{378622}{1551312}\ . \end{align}