The probability that $S_n = k$ is of course $\binom{n}{k}p^k(1-p)^{n-k}$.
For $X_1$ to be $1$ and also $S_n = k$ two things must be true: $X_1 = 1$ (probability $p$) and $k-1$ among the remaining $n-1$ $X$-es must be $1$ (probability $\binom{n-1}{k-1}p^{k-1}(1-p)^{n-1-(k-1)}$).
So the conditional probability you want is
$$
\frac{p \binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}}{\binom{n}{k}p^k(1-p)^{n-k}}=
\frac{\binom{n-1}{k-1}}{\binom{n}{k}}=\frac{(n-1)!}{(k-1)!(n-k)!}\frac{k!(n-k)!}{n!} = \frac{k}{n}
$$
Now you can slap your forhead, because it was trivial to see that $k$ out of the $n$ variables were $1$, so the probability of any specific variable being $1$ would have to be $\frac{k}{n}$. All that work for such an easy result!
As you noticed, $S_n$ is a sum of $n-$ independent copies of random variable $X$ such that $\mathbb P(X=1)=p = 1-\mathbb P(X=0)$, so it's the case $S_n \sim \mathcal B(n,p)$.
For $T_n$, notice that the random variable $Z_k = X_k Y_k$ is also $\{0,1\}$ random variable and due to independence of $X_k,Y_k$ we have $\mathbb P(Z_k = 1) = \mathbb P(X_k=1, Y_k = 1) = pq$ and $\mathbb P(Z_k = 0) = 1 -pq$.
That means: $T_n$ is also a sum of $n-$ independent copies of random variable $Z$ such that $\mathbb P(Z=1) = pq = 1- \mathbb P(Z=0)$, so $T_n \sim \mathcal B(n,pq)$.
For $N$, notice that $\inf\{n \ge 0 : T_{n+1} = 1\} = \inf \{n \ge 0 : Z_{n+1} = 1\}$
So to have $\{N=k\}$ for some $k \in \{0,1,2,...\}$ we need to have $Z_{1},...,Z_{k}$ being equal to $0$ and $Z_{k+1}$ being equal to $1$.
That is $\mathbb P(N=k) = \mathbb P(Z_1=0,...,Z_k = 0, Z_{k+1} = 1) = \mathbb P(Z=0)^k \mathbb P(Z=1) = (1-pq)^k \cdot pq$
Equalities due to independence and the same distribution of $Z_k$ as random variable $Z$ (defined above)
So $N \sim Geo(pq)$
Best Answer
The conditional distribution you are looking for is the following
$$ P(X_i=k) = \begin{cases} \frac{n-r}{n}, & \text{if k=0} \\ \frac{r}{n}, & \text{if k=1} \end{cases}$$
it is very easy to prove..
Without loss of generality let's prove the conditional distribution of $X_1$ (all the variables have the same density)
$$\mathbb{P}[X_1=0|\sum_iX_i=r]=\frac{\mathbb{P}[X_1=0;\sum_{i=1}^nX_i=r]}{\mathbb{P}[\sum_{i=1}^nX_i=r]}=$$
$$=\frac{\mathbb{P}[X_1=0;\sum_{i=2}^nX_i=r]}{\mathbb{P}[\sum_{i=1}^nX_i=r]}=\frac{\mathbb{P}[X_1=0]\mathbb{P}[\sum_{i=2}^nX_i=r]}{\mathbb{P}[\sum_{i=1}^nX_i=r]}=$$
$$\frac{(1-p)\binom{n-1}{r}p^r(1-p)^{n-1-r}}{\binom{n}{r}p^r(1-p)^{n-r}}=\frac{(n-1)!r!(n-r)!}{r!(n-1-r)!n!}=\frac{n-r}{n}$$
Similar brainstorming for $\mathbb{P}[X_1=1]$