a) We solve this problem by first finding the probability of NONE of the $17$ people having the same birthday.
Person $1$ has a birthday (let's call it $D_1$). Then Person $2$ must have a birthday other than $D_1$ (let's call it $D_2$), which leaves him $365-1 = 364$ options. Person $3$ must have a birthday other than $D_1$ and $D_2$ (let's call it $D_3$), which leaves him $365-2 = 363$ options. This goes on until you reach $D_{17}$.
Because there are $365^{17}$ arrangements of birthday possible, and you have $365 \cdot 364 \cdot 363 \cdots 349$ options, the answer is $\displaystyle \frac{365 \cdot 364 \cdot 363 \cdots 349}{365^{17}}$.
Note that we're not finished yet, however, because we found the probability that NONE of them have the same birthday when we really want the probability that AT LEAST TWO have the same birthday. Fortunately, the math is simple - we just subtract the value we found from $1$. So the final answer is $$ 1 - \frac{365 \cdot 364 \cdot 363 \cdots 349}{365^{17}}.$$
b) We can choose two people among the $17$ to have the birthday of January $1$st, and we don't care when the others have their birthdays as long as it's not January $1$st.
We have to choose the two people. Because order doesn't matter, this will be a combination - namely, $\dbinom{17}{2}$. The probability that both of them have the birthday of January $1$st is $\displaystyle \left(\frac{1}{365}\right)^2$.
We then have to assign a birthday to each of the others (N.B.: They can have the same birthdays), which means the probability will be $\displaystyle \left(\frac{364}{365}\right)^{15}$).
Multiplying all of these together, we get
$$\dbinom{17}{2} \left(\frac{1}{365}\right)^2 \left(\frac{364}{365}\right)^{15}.$$
As far as I can tell, your methods do not work and cannot be salvaged.
This is a hard problem that cannot be solved by conventional means. To find the probability that at least three people have a common birthday, let us instead compute the probability that no three people share a birthday. This is equal to
$$
\frac{n^{365}}{365^n}.\tag 1
$$
Here, $[x^n]f(x)$ is the coefficient of $x^n$ in the polynomial $f(x)$. You want one minus the above. Another way to write this is
$$
\sum_{a_1,a_2,\dots,a_{365}}\frac1{365^n}\frac{n!}{a_1!a_2!\cdots a_{365}!}.\tag 2
$$
where the sum ranges over all vectors $(a_1,a_2,\dots,a_{365})$ of integers between $0$ and $2$ whose sum is $n$. This works because each vector specifies a valid distribution of birthdays where no birthday repeats three times or more, and the multinomial coefficient gives the number of ordered selections of birthdays which have that distribution, and each is then multiplied by the probability $(1/365)^n$ of that ordered selection occurring. You can check that when you expand out $(1)$ and collect the coefficient of $x^{n}$, you get exactly $(2)$.
If you are okay with an approximate result, the expected number of triplets of people who share a common birthday is $\lambda =\binom{n}{3}\cdot \frac1{365}^2$, and the number of triplets with a common birthday is approximately Poisson with parameter $\lambda$. Therefore, the probability some triplet share a birthday is approximately
$$
1-e^{-\binom{n}3/365^2}.
$$
Best Answer
Given $2k$ items, there are $(2k-1)!!$ ways to arrange them into pairs: the first item can be paired with $2k-1$ possibilities; the first unpaired item can be matched with $2k-3$ items; the new first unpaired item can be paired with $2k-5$ items; etc.
The number of functions from $n$ people to $365$ dates with $n-2k$ singles and $k$ pairs is $$ \begin{array}{cc} &\displaystyle\underbrace{ \overbrace{\binom{365}{n-k}}^{\substack{\text{ways to choose}\\\text{$n-k$ dates}\\\text{for birthdays}}} \overbrace{\binom{n-k}{k}}^{\substack{\text{ways to choose}\\\text{$k$ dates}\\\text{for pairs}}} }&\displaystyle\underbrace{ \overbrace{\ \ \binom{n}{2k}\ \ }^{\substack{\text{ways to choose}\\\text{$2k$ people}\\\text{for pairs}}} \overbrace{(n-2k)!\vphantom{\binom{n}{k}}}^{\substack{\text{ways to arrange}\\\text{$n-2k$ singles}}} \overbrace{(2k-1)!!\vphantom{\binom{n}{k}}}^{\substack{\text{ways to pair}\\\text{$2k$ people}}} \overbrace{\ \ \ \ \ k!\ \ \ \ \ \vphantom{\binom{n}{k}}}^{\substack{\text{ways to arrange}\\\text{$k$ pairs}}} }\\ \displaystyle= &\displaystyle\frac{365!}{(365-n+k)!\,(n-2k)!\,k!} &\displaystyle\frac{n!}{2^k} \end{array} $$ Thus, the probability of getting at least one triple is $$ 1-\frac{365!\,n!}{365^n}\sum_{k=0}^{365}\frac1{(365-n+k)!\,(n-2k)!\,k!\,2^k} $$ where we take $\frac1{n!}=0$ for $n\lt0$.
Here is a plot from $n=0$ to $n=730$. For $n\lt3$, the probability of getting a triple is $0$, and for $n\gt730$, the probability is $1$.