1. Solve the system of congruences: (1) $\left\{\begin{array}{l}x \equiv 3(\bmod 7) \\ 6 x \equiv 10(\bmod 8)\end{array}\right.$
(1)
(2) $\left\{\begin{array}{l}3 x \equiv 1(\bmod 10) \\ 4 x \equiv 7(\bmod 15)\end{array}\right.$
For the system of equations (I), first solve (2), we get $x \equiv -1, 3 (\bmod 8)$, so the solution to (1) is the solution to the following two systems of congruences: $x \equiv 3 (\bmod 7), x \equiv -1 (\bmod 8)$, or $x \equiv 3 (\bmod 7), x \equiv 3 (\bmod 8)$.
Using the Chinese Remainder Theorem, the solution to (I) is $x \equiv 3, 31 (\bmod 56)$.
For the system of equations (II), it is equivalent to the following system of congruences:
$$
\left\{\begin{array}{l}
3 x \equiv 1(\bmod 2), \\
3 x \equiv 1(\bmod 5), \\
4 x \equiv 7(\bmod 3), \\
4 x \equiv 7(\bmod 5) .
\end{array}\right.
$$

But from $3 x \equiv 1(\bmod 5) \Rightarrow x \equiv 2(\bmod 5)$,
and from $4 x \equiv 7(\bmod 5) \Rightarrow x \equiv 3(\bmod 5)$, which is a contradiction!
Therefore, the system of equations (II) has no solution.