We may as well suppose the game continues until all targets have been hit (which will happen eventually if all $p_j > 0$; we may as well remove any targets that have $p_j = 0$).
For each subset $S$ of $\{1,\ldots, N\}$, let $p_S = \sum_{s \in S} p_s$ be the probability that a shot hits a member of $S$, and let
$a_{i,t}(S)$ be the probability that target $i$ is one of the first $t$ targets in set $S$ to be hit. You want $1 - a_{i,X}(\{1,\ldots,N\})$.
Of course $a_{i,t}(S) = 0$ if $i \notin S$, and we also take it to be $0$ if
$t = 0$.
Otherwise, conditioning on the first target in $S$ to be hit,
$$ a_{i,t}(S) = \dfrac{p_i}{p_S} + \sum_{j \in S \backslash \{i\}} \dfrac{p_j}{p_S} a_{i,t-1}(S \backslash \{j\}) $$
Now I claim that
$$ a_{i,t}(S) = \sum_{T \subseteq S \backslash \{i\}} c(t,|T|,|S|) \frac{p_i}{p_{T \cup \{i\}}} $$
for some constants $c(t,m,n)$, $0 \le m \le n-1$.
I will prove this by induction on $t$.
In the case $t=1$ we have $a_{i,1}(S) = p_i/p_S$, so $c(1,m,n) = 1$ if
$m = n-1$, $0$ otherwise.
If $t >1$, we have (with $|S|=n$):
$$ \eqalign{a_{i,t}(S) &= \dfrac{p_i}{p_S} + \sum_{j \in S \backslash \{i\}}
\dfrac{p_j}{p_S}
\sum_{T \subseteq S \backslash \{i,j\}} c(t-1, |T|,n-1) \dfrac{p_i}{p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \sum_{j \in S \backslash (T \cup \{i\})} \dfrac{p_j p_i}{p_S p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \dfrac{p_{S \backslash (T \cup \{i\})} p_i}{p_S p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \dfrac{(p_S - p_{T \cup \{i\}}) p_i}{p_S p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \left(\dfrac{p_i}{p_{T \cup \{i\}}} - \dfrac{p_i}{p_S}\right)\cr
&= \sum_{T \subseteq S \backslash \{i\}} c(t,|T|,n) \dfrac{p_i}{p_{T \cup \{i\}}}
}$$
where $c(t, m,n) = c(t-1, m,n-1)$ if $m < n-1$ while
$$c(t, n-1, n) = 1 - \sum_{m=0}^{n-2} {n-1 \choose m} c(t-1,m,n-1)$$
Hmm: it looks like
$$ c(t, m, n) = \cases{1 & for $t=n,m=0$\cr
(-1)^{n+m+t} {m-1 \choose {t+m-n}} & $ n-m \le t \le n$\cr
0 & otherwise\cr}$$
There ought to be an inclusion-exclusion proof for this.
Your solution to the first problem is correct.
Let $A$ be the event that the first shooter hits the target; let $B$ be the event that the second shooter hits the target; let $C$ be the event that the third shooter hits the target. If at least two shooters hit the target, then either exactly two of them hit the target or all three do. Thus, we must calculate
$$\Pr(A)\Pr(B)\Pr(C^C) + \Pr(A)\Pr(B^C)\Pr(C) + \Pr(A^C)\Pr(B)\Pr(C) + \Pr(A)\Pr(B)\Pr(C)$$
given $\Pr(A) = 0.70$, $\Pr(B) = 0.80$, $\Pr(C) = 0.90$. Can you proceed?
Best Answer
EDIT:I have included an explicit answer for the scenario that both players' first shots are fired simultaneously below.
Martigan's answer above works well for the scenario that both player's first shots coincide. Because it is an interesting problem, here below is the other situation, if their shots are offset.
This can be approached via markov chains if they never fire at the same time (i.e. player A's first shot does not coincide with either player B's first shot or player B's second shot). I will make the additional assumption that neither player gets a substantial head start.
This can then be modeled with a Markov Chain with the following transition diagram:
Represented by the matrix with order $A_w, B_w, A, B_1, B_2$ as
$\begin{bmatrix} 1 & 0 & .8 & 0 & 0\\ 0 & 1 & 0 & .5 & .5\\ 0 & 0 & 0 & 0 & .5\\ 0 & 0 & .2 & 0 & 0\\ 0 & 0 & 0 & .5 & 0\end{bmatrix}$
This is in the form $A=\begin{bmatrix} I & S\\ 0 & R\end{bmatrix}$ which $\lim_{n\to\infty} A^n = \begin{bmatrix}I&S(I-R)^{-1}\\0 & 0\end{bmatrix}$
Through matrix arithmetic, we get that $(I-R)^{-1} \approx \begin{bmatrix}1.05 & 0.26 & 0.53\\ 0.21 & 1.05 & 0.11\\ 0.11 & 0.53 & 1.05\end{bmatrix}$
and that $\lim_{n\to\infty} A^n = \begin{bmatrix}1 & 0 & 0.84 & 0.21 & 0.42\\ 0 & 1 & 0.16 & 0.79 & 0.58\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\end{bmatrix}$
Thus, if player A shoots first, player A wins with probability 0.84
If player B has two shots before player A, player B wins with probability 0.79
If player B has only one shot before player A, player B wins with probability 0.58
Because of the confusion among other posters, I decided to explicitly solve the interpretation that both snipers' first shots are at the exact same time and there is no tie allowed (if they both hit, they both shoot again).
The transition diagram then is
The transition matrix with order $A_w, B_w, AB_1, B_2$ is: $\begin{bmatrix} 1 & 0 & .4 & 0\\ 0 & 1 & .1 & .5\\ 0 & 0 & 0 & .5\\ 0 & 0 & .5 & 0\end{bmatrix}$
Fundamental matrix then is $\begin{bmatrix}1 & -.5 \\ -.5 & 1\end{bmatrix}^{-1} = \begin{bmatrix}4/3 & 2/3\\2/3&4/3\end{bmatrix}$
And the limiting matrix becomes $\begin{bmatrix}1 & 0 & .5\overline{3} & .2\overline{6}\\ 0 & 1 & .4\overline{6} & .7\overline{3}\\ 0& 0 & 0 & 0\\ 0&0&0&0\end{bmatrix}$
Thus, the chance that player $A$ wins is $0.5\overline{3}$ assuming they both start at the same time, and therefore player $A$ is more likely to win.