For the first conditional probability, we can calculate. We have
$$\Pr(T=k+m|T\gt k)=\frac{\Pr(T=k+m \cap T\gt k)}{\Pr(T\gt k)}.$$
The numerator is $(1-p)^{m+k-1}p$, and the denominator is $(1-p)^k$. Divide. We get $(1-p)^{m-1}p$.
The above result is intuitively clear. It captures the memorylessness of the distribution. The coin does not remember its history of failures.
The above calculation can serve as a template for the second problem.
Added: Again, this is a conditional probability calculation. For the numerator, we want $\Pr(T_1=k \cap T_1+T_2=n)$, or equivalently $\Pr(T_1=k \cap T_2=n-k)$. By independence, this is $(1-p)^{k-1}p(1-p)^{n-k-1}p$, which simplifies to $(1-p)^{n-2}p^2$.
For the denominator, we want the probability that $T_1+T_2=n$. This event happens if $T_1=1$ and $T_2=n-1$ or if $T_1=2$ and $T_2=n-2$, and so on.
The probability that $T_1=i$ and $T_2=n-i$ is, by a calculation identical to the one above, equal to $(1-p)^{n-2}p^2$. For the probability that $T_1+T_2=n$, we add from $i=1$ to $i=n-1$. So we get $(n-1)(1-p)^{n-2}p^2$.
Finally, for the conditional probability, divide. We get the constant $\frac{1}{n-1}$. This says that the conditional distribution of $T_1$, given the sum $T_1+T_2$ is $n$, is (discrete) uniform.
The joint distribution of the random variables $X$ and $P$ is given by
$$pr(X=x,P=p) = pr(X=x\ |\ P=p) \ pr(P=p),$$
where
$$pr(X=x\ |\ P=p) = \binom{6}{x}p^x (1-p)^{6-x}\ (x=0,1,2,3,4,5,6)
$$
and $pr(P=p)$ is given by your table, for $p=0.2, 0.6, 0.8$.
The unconditional mean and variance of $X$ is found from the marginal distribution
$$pr(X=x) = \sum_{p} pr(X=x,P=p).
$$
This should be enough to answer all of the stated questions.
Best Answer
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!