381. The numbers $p$ and $p+15$ are prime. Find all such $p$.
$\triangle$ When $p=2$, we get that the number $p+15=17$ is prime.

Let $p>2$. Then $p$ is odd. Consequently, the number $p+15$ is even. Hence, this number is composite. Therefore, no prime $p>2$ works.

Answer: 2.

$382^{\circ}$. Prove that the numbers $p, p+2$, and $p+4$ are all prime only in the case when they form the triplet $3,5,7$.

$\triangle$ Consider several cases depending on $p$.

When $p=2$, the number $p+2=4$ is composite, so the value $p=2$ is ruled out.

When $p=3$, we get the triplet $3,5,7$, mentioned in the problem statement.

When $p=5$, the number $p+2=7$ is prime, but the number $p+4=9$ is composite, so $p=5$ must be discarded.

When $p=7$, the number $p+2=9$ is composite.

When $p=11$, the number $p+4=15$ is also composite.

It seems that only $p=3$ works. Let's prove it.

It is not hard to notice that the values $p=5, p=7$, and $p=11$ did not work because either $p+2$ or $p+4$ is divisible by 3. Let's confirm that this will always be the case for a prime $p>3$.

A prime number greater than 3 does not divide by 3 and, therefore, when divided by 3, can only leave a remainder of 1 or 2. Let's consider both cases.

1) Suppose $p$ when divided by 3 leaves a remainder of $1: p=3k+1(k \in \mathbb{N})$. Then the number

$$
p+2=(3k+1)+2=3k+3
$$

is divisible by 3, and the quotient of this division is greater than 1. Thus, the number $p+2$ is composite.

2) Suppose $p=3k+2(k \in \mathbb{N})$. Then the number

$$
p+4=(3k+2)+4=3k+6
$$

is composite.