Yes.
You could think of each coin having a three-state outcome, with $P[\mathrm{Heads}]=p_i-x_i$, of $P[\mathrm{Tails}]=1-p_i$ and $P[\mathrm{Other}]=x_i$, where the Other event could be either with some distribution determined by your previous tosses.
Then your probability of all Heads is at least $\prod (p_i-x_i)$, and your probability of at least one Tails is at least $1-\prod p_i$, establishing your bounds.
Let $A_{m,n}$ be the probability that player $A$ flips two consecutive heads before player $B$ flips three consecutive heads, given that $A$'s (resp. $B$'s) current run of heads has length $m$ (resp. $n$). Then
$$
A_{m,n}=\frac{1}{4}\left(A_{m+1,n+1} + A_{m+1,0} + A_{0,n+1}+A_{0,0}\right);
$$
the boundary conditions are that $A_{m,n}=1$ for $m\ge 2$ and $n < 3$, and $A_{m,n}=0$ for $m \le 2$ and $n\ge 3$. You want to find $A_{0,0}$. The relevant six equations are:
$$
\begin{eqnarray}
A_{0,0} &=& \frac{1}{4}A_{1,1} + \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,1} + \frac{1}{4}A_{0,0}\\
A_{0,1} &=& \frac{1}{4}A_{1,2} + \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,2} + \frac{1}{4}A_{0,0} \\
A_{1,0} &=& \frac{1}{2} + \frac{1}{4}A_{0,1} + \frac{1}{4}A_{0,0} \\
A_{1,1} &=& \frac{1}{2} + \frac{1}{4}A_{0,2} + \frac{1}{4}A_{0,0} \\
A_{0,2} &=& \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,0} \\
A_{1,2} &=& \frac{1}{4} + \frac{1}{4}A_{0,0},
\end{eqnarray}
$$
or
$$
\left(\begin{matrix}3/4 & -1/4 & -1/4 & -1/4 & 0 & 0 \\
-1/4 & 1 & -1/4 & 0 & -1/4 & -1/4 \\
-1/4 & -1/4 & 1 & 0 & 0 & 0 \\
-1/4 & 0 & 0 & 1 & -1/4 & 0 \\
-1/4 & 0 & -1/4 & 0 & 1 & 0 \\
-1/4 & 0 & 0 & 0 & 0 & 1
\end{matrix}\right)\times\left(\begin{matrix} A_{0,0} \\ A_{0,1} \\ A_{1,0} \\ A_{1,1} \\ A_{0,2} \\ A_{1,2}
\end{matrix}\right)
= \left(\begin{matrix}
0 \\
0 \\
1/2 \\
1/2 \\
0 \\
1/4
\end{matrix}\right),
$$
assuming no typos. Further assuming no typos entering this into WolframAlpha, the result is
$$
A_{0,0} = \frac{1257}{1699} \approx 0.7398,
$$
which at least looks reasonable.
Update: As pointed out in a comment, the above calculation finds the probability that $A$ gets two consecutive heads sooner than $B$ gets three consecutive heads; the original problem asks for the opposite. The correct boundary conditions for the original problem are that $A_{m,n}=1$ for $m<2$ and $n\ge 3$ and $A_{m,n}=0$ for $m\ge 2$ and $n\le 3$. The matrix equation becomes
$$
\left(\begin{matrix}3/4 & -1/4 & -1/4 & -1/4 & 0 & 0 \\
-1/4 & 1 & -1/4 & 0 & -1/4 & -1/4 \\
-1/4 & -1/4 & 1 & 0 & 0 & 0 \\
-1/4 & 0 & 0 & 1 & -1/4 & 0 \\
-1/4 & 0 & -1/4 & 0 & 1 & 0 \\
-1/4 & 0 & 0 & 0 & 0 & 1
\end{matrix}\right)\times\left(\begin{matrix} A_{0,0} \\ A_{0,1} \\ A_{1,0} \\ A_{1,1} \\ A_{0,2} \\ A_{1,2}
\end{matrix}\right)
= \left(\begin{matrix}
0 \\
0 \\
0 \\
0 \\
1/2 \\
1/4
\end{matrix}\right),
$$
The result becomes $A_{0,0}=\frac{361}{1699}\approx 0.2125$. The two results add to slightly less than one because there is a nonzero probability that both players hit their goals at the same time... this probability is $81/1699\approx 0.0477$.
Best Answer
With $\mathbb{P}[X=i]$ and $\mathbb{P}[Y=i]$ figured out for $i\in\{0,1,2,3\}$ you are done, since $\mathbb{P}[X=i,Y=j]=\mathbb{P}[X=i]\mathbb{P}[Y=j]$ for any $i\in\{0,1,2,3\}$ and $j\in\{0,1,2,3\}$ since the two coin tosses are independent.
$\mathbb{P}[X>Y]$ is just sum of $\mathbb{P}[X=i,Y=j]$ over $i\in\{0,1,2,3\}$ and $j\in\{0,1,2,3\}$ such that $i>j$.