4. Let's write the natural number $a$ in its canonical form:

$$
a=p_{1}^{s_{1}} \cdot p_{2}^{s_{2}} \cdot \ldots \cdot p_{n}^{s_{n}}
$$

where $p_{1}, p_{2}, \ldots, p_{n}$ are distinct prime numbers, and $s_{1}, s_{2}, \ldots, s_{n}$ are natural numbers.

It is known that the number of natural divisors of $a$, including 1 and $a$, is equal to $\left(s_{1}+1\right)\left(s_{2}+1\right) \cdot \ldots \cdot\left(s_{n}+1\right)$. According to the problem, this number is 101. Since 101 is a prime number, all the brackets in the product are equal to 1, except for one which is equal to 101. Therefore, the number $a=p^{100}$, and its divisors are $d_{1}=1, d_{2}=p, d_{3}=p^{2}, \ldots$, $d_{101}=p^{100}$.

Let's calculate the sum of the divisors:

$$
\Sigma_{d}=1+p+p^{2}+\ldots+p^{100}=\frac{p^{101}-1}{p-1}=\frac{a \sqrt[100]{a}-1}{\sqrt[100]{a}-1}
$$

Let's calculate the product of the divisors:

$$
\Pi_{d}=p^{1+2+\ldots+100}=p^{50 \cdot 101}=\sqrt{a^{101}}
$$
Answer: 1) $\Sigma_{d}=\frac{a_{100}^{a}-1}{\sqrt[100]{a}-1}$; 2) $\Pi_{d}=\sqrt{a^{101}}$.