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 $x \equiv -1, 3 \pmod{8}$, so the solutions to (1) are the solutions to the following two systems of congruences: $x \equiv 3 \pmod{7}, x \equiv -1 \pmod{8}$, or $x \equiv 3 \pmod{7}, x \equiv 3 \pmod{8}$.

Using the Chinese Remainder Theorem, the solutions to (I) are $x \equiv 3, 31 \pmod{56}$.
For the system of equations (II), it is equivalent to the following system of congruences:
$$\left\{\begin{array}{l}
3 x \equiv 1 \pmod{2}, \\
3 x \equiv 1 \pmod{5}, \\
4 x \equiv 7 \pmod{3}, \\
4 x \equiv 7 \pmod{5}.
\end{array}\right.$$

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