3. Let $p$ be an odd prime, and $s(M)$ denote the sum of all elements in set $M$. For the non-empty subsets $T$ of set $F=\{1,2, \cdots, p-1\}$, if $p \mid s(T)$, find the number of such $T$.


3. The following proof: There are $\frac{2^{p-1}-1}{p}+1$ subsets that satisfy the condition, then there are $\frac{2^{p-1}-1}{p}$ non-empty subsets $T$ that satisfy the condition. Let $F_{1}=\{1,2, \cdots, p\}$.
Since each subset $A$ of $F$ corresponds to two subsets $A$ and $A \cup\{p\}$ in $F_{1}$, the number of subsets of $F_{1}$ whose elements sum to a multiple of $p$ is twice the number of subsets of $F$.

The following proof: There are $\frac{2^{p}-2}{p}+2$ subsets of $F_{1}$ whose elements sum to a multiple of $p$.

Since $s\left(F_{1}\right)$ and $s(\varnothing)$ are both divisible by $p$, it suffices to prove that the number of subsets of $F_{1}$ with elements in the range $1 \sim p-1$ and whose elements sum to a multiple of $p$ is $\frac{2^{p}-2}{p}$.
For two subsets of $F_{1}$:
$$
\begin{array}{l}
A=\left\{a_{1}, a_{2}, \cdots, a_{i}\right\}, \\
B=\left\{b_{1}, b_{2}, \cdots, b_{i}\right\},
\end{array}
$$

if $b_{j} \equiv a_{j}+m(\bmod p)(1 \leqslant j \leqslant i, m \neq 0(\bmod p))$, then $A$ and $B$ are called "good sets".

Since $m$ modulo $p$ can take any non-zero integer, $A$ has $p-1$ good sets.
For $i(i \neq 0$, and $i \neq p)$, by
$$
s(B) \equiv s(A)+i m \not \equiv s(A)(\bmod p),
$$

we know that these good sets $B$ are all different from $A$.
If sets $A$, $B$, and $C$ are different, and $B$ and $C$ are good sets with $A$, then $B$ and $C$ are good sets with each other.

In particular, if $A$ and $B$ are good sets, then $B$ is also a good set with the other $p-2$ good sets of $A$.

Since each set has at most $p-1$ good sets, these sets are all the good sets of $B$.
Let $I(A)$ be the union of $A$ and its good sets.
It has been proven: If $A$ and $B$ are good sets, then
$I(A)=I(B)$.
Also, if $I(A)$ and $I(B)$ have a common good set $C$, then $A$ and $B$ must be good sets.
Thus, if $A$ and $B$ are not good sets, then
$I(A) \neq I(B)$.
Assume that the $p$ sets in $I(A)$ have elements whose sums modulo $p$ are all different. Then exactly one set has an element sum that is divisible by $p$.

Therefore, among the subsets with elements in the range $1 \sim p-1$, exactly one is in $I(A)$.

Since the number of subsets of $F_{1}$ is $2^{p}$, the number of subsets with elements in the range $1 \sim p-1$ is $2^{p}-2$.

Thus, the number of subsets of $F_{1}$ with elements in the range $1 \sim p-1$ and whose elements sum to a multiple of $p$ is $\frac{2^{p}-2}{p}$.