I am considering the ring $R=\mathbb{Z}[\frac{1}{5}]=\{p(\frac{1}{5})\mid p(x) \in \mathbb{Z}[x]\}$ and I want to determine whether the ideals $I=\left \langle 2,\frac{1}{5}\right \rangle$ and $J=\left \langle 2,\frac{3}{5}\right \rangle$ are principal ideals.
So far, I think that $I=\left \langle \frac{1}{5}\right \rangle$ and is hence principal. Clearly, we have $\left \langle \frac{1}{5}\right \rangle \subset I$. To show the other direction, observe that if $\alpha=2p(\frac{1}{5})+\frac{1}{5}q(\frac{1}{5})\in I$ for some $p,q \in \mathbb{Z}[x]$, we can write $\alpha = \frac{1}{5}\left(10p(\frac{1}{5})+q(\frac{1}{5})\right)\in \left \langle \frac{1}{5}\right \rangle$. Hence $I \subset \left \langle \frac{1}{5} \right \rangle \implies I =\left \langle \frac{1}{5} \right \rangle$.
For the second case, I believe that $J=\left \langle 2, \frac{3}{5} \right \rangle$ is not principal in $R$. To prove this, I have tried assuming that $J=\left \langle t\left( \frac{1}{5} \right) \right \rangle$ for some $t(x)\in\mathbb{Z}[x]$. Then clearly, we must have that $t\left( \frac{1}{5} \right)$ divides $2$ and $\frac{3}{5}$. But I am struggling to place any restrictions on $t$: for instance, if we assume that $\frac{3}{5}=t\left(\frac{1}{5}\right)r\left(\frac{1}{5}\right)$ with $r \in \mathbb{Z}[x]$, then I don't believe I can say much about the degree of $t$ and $r$, since it is possible the coefficients will cancel. As an example, I can write $\frac{3}{5}=(3\cdot 5)\frac{1}{5^2}=(3\cdot 5^2)\frac{1}{5^3}$ and so on... As a result, I am at a loss for how to show that $J$ is or is not principal in $R$.
I imagine that I am making things too complicated, or have not understood the definition of $R$.
Two facts can be used to deal with both of these.
(1) If an ideal $I$ of a ring $R$ contains a unit, then $I = \langle 1 \rangle = R$.
(2) Given $r \in R$ and a unit $u \in R$, then $\langle r \rangle = \langle ur \rangle$, i.e., associates (elements that differ multiplicatively by a unit) generate the same ideal.
Returning to your problem, since $1/5$ is a unit then (1) immediately shows that $I = \langle 1 \rangle$. Again, since $1/5$ is a unit then (2) implies $\langle 2, 3/5 \rangle = \langle 2, 3 \rangle$. Since $1 = 3 - 2 \in \langle 2, 3 \rangle$, this shows that $J = \langle 1 \rangle$ as well.