a) Let $p$ be a prime and suppose that $p$ is a divisor of $D_{n}$. Then $p$ is a divisor of

$$\left((a+1)^{n}+(a+2)^{n}+(a+3)^{n}\right)-\left(a^{n}+(a+1)^{n}+(a+2)^{n}\right)=(a+3)^{n}-a^{n}$$

for all positive integers $a$. Choose now $a=p$, then $p \mid(p+3)^{n}-p^{n}$, or $(p+3)^{n}-p^{n} \equiv 0 \bmod p$. This is simply $3 \equiv 0 \bmod p$, so $p=3$. We conclude that $D_{n}$ only contains prime factors 3 and thus is of the form $3^{k}$ with $k \geq 0$.
b) For $k=0$ we take $n=2$. It holds that $1^{2}+2^{2}+3^{2}=14$ and $2^{2}+3^{2}+4^{2}=29$ and these two have no prime factors in common, so $D_{2}=1$. Now assume that $k \geq 1$. We will prove that $D_{n}=3^{k}$ for $n=3^{k-1}$.
First, we show that $1^{n}+2^{n}+3^{n}$ for $n=3^{k-1}$ is divisible by $3^{k}$, but not by $3^{k+1}$. For $k=1$ it is $n=1$ and indeed $1+2+3=6$ is divisible by 3, but not by $3^{2}$. For $k \geq 2$ it holds that $n>k$ and thus $3^{n}$ is divisible by $3^{k+1}$. What we need to prove is thus equivalent to: $1+2^{n}$ for $n=3^{k-1}$ is divisible by $3^{k}$ but not by $3^{k+1}$. We prove this by induction on $k$. For $k=2$ it is $n=3$ and indeed $1+8=9$ is divisible by 9, but not by 27. Let $m \geq 2$ and assume we have proven this for $k=m$. Take $n=3^{m-1}$. We know that $1+2^{n}$ is divisible by $3^{m}$, but not by $3^{m+1}$. We want to show that $1+2^{3 n}$ is divisible by $3^{m+1}$, but not by $3^{m+2}$. Write $1+2^{n}=3^{m} c$ with $3 \nmid c$. Then $2^{n}=3^{m} c-1$, so

$$1+2^{3 n}=1+\left(3^{m} c-1\right)^{3}=3^{3 m} c^{3}-3 \cdot 3^{2 m} c^{2}+3 \cdot 3^{m} c$$

Modulo $3^{m+2}$ this is congruent to $3^{m+1} c$ and since $3 \nmid c$ it follows that this is divisible by $3^{m+1}$, but not by $3^{m+2}$, as we wanted to prove. This completes the induction. Now we show that for $n=3^{k-1}$ it holds that $(a+3)^{n}-a^{n}$ is divisible by $3^{k}$ for all positive integers $a$. We prove this again by induction on $k$. For $k=1$ it is $n=1$ and indeed $(a+3)-a=3$ is divisible by 3. Now let $m \geq 1$ and assume we have proven this for $k=m$. Take $n=3^{m-1}$. Then we know that $(a+3)^{n}-a^{n}$ is divisible by $3^{m}$, so we can write $(a+3)^{n}=a^{n}+3^{m} c$ for some integer $c$. Taking the third power on both sides gives

$$(a+3)^{3 n}=a^{3 n}+3 a^{2 n} \cdot 3^{m} c+3 a^{n} \cdot 3^{2 m} c^{2}+3^{3 m} c^{3},$$

so

$$(a+3)^{3 n}-a^{3 n}=a^{2 n} \cdot 3^{m+1} c+a^{n} \cdot 3^{2 m+1} c^{2}+3^{3 m} c^{3} .$$

This is divisible by $3^{m+1}$, which completes the induction.
We have now proven for $n=3^{k-1}$ that $3^{k} \mid 1^{n}+2^{n}+3^{n}$ and $3^{k} \mid(a+3)^{n}-a^{n}$ for all positive integers $a$, from which it follows directly by induction on $a$ that $3^{k} \mid a^{n}+(a+1)^{n}+(a+2)^{n}$ for all $a$. Thus $3^{k} \mid D_{n}$. Since $3^{k+1} \nmid 1^{n}+2^{n}+3^{n}$ it also holds that $3^{k+1} \nmid D_{n}$. Therefore, $D_{n}=3^{k}$.