[Solution] (1) If $n$ is a multiple of 3, then we can set $n=3k$ (where $k$ is a positive integer),
$$\begin{aligned} 2^{n}-1 & =2^{3k}-1 \\ & =8^{k}-1 \\ & =(8-1)\left(8^{k-1}+8^{k-2}+\cdots+1\right) \end{aligned}$$

Therefore, $2^{n}-1$ is divisible by 7.
If $n$ is not a multiple of 3, then we can set $n=3k+1, n=3k+2$ (where $k$ is a non-negative integer).
$$\begin{aligned} \text { When } n=3k & +1 \text {, } \\ 2^{n}-1 & =2^{3k+1}-1 \\ & =2 \cdot 2^{3k}-1 \\ & =2 \cdot 8^{k}-1 \\ & =2(7+1)^{k}-1 . \end{aligned}$$

Since $(7+1)^{k}$ leaves a remainder of 1 when divided by 7, $2(7+1)^{k}$ leaves a remainder of 2 when divided by 7, and $2(7+1)^{k}-1$ leaves a remainder of 1 when divided by 7. Thus, $2^{n}-1$ leaves a remainder of 1 when divided by 7.

When $n=3k+2$, $2^{n}-1$
$$\begin{array}{l} =4 \cdot 8^{k}-1 \\ =4 \cdot(7+1)^{k}-1 \end{array}$$

Since $4 \cdot(7+1)^{k}$ leaves a remainder of 4 when divided by 7, $4 \cdot(7+1)^{k}-1$ leaves a remainder of 3 when divided by 7. Thus, $2^{n}-1$ leaves a remainder of 3 when divided by 7.

In summary, $2^{n}-1$ is divisible by 7 if and only if $n$ is a multiple of 3.
(2) From (1), we know that when $n=3k, 3k+1, 3k+2$, the remainders of $2^{n}$ when divided by 7 are $1, 2, 4$ respectively. Therefore, the remainders of $2^{n}+1$ when divided by 7 are $2, 3, 5$ respectively. This means that for any positive integer $n$, $2^{n}+1$ is never divisible by 7.