Algebraic number theory, Marcus, Chapter 3, Question 9

It's not terrible, just prove ($2$) and then note $IS|JS \iff JS\subseteq IS$, now intersect both sides with $R$ and you get $J= R\cap SJ\subseteq R\cap IS = I$.

For ($2$) you get the intersection properties come to you since you can do it for prime powers, and then using the fact that distinct prime powers' intersections are their product.

For the last one, it's easy to see based on how primes split in the extension, so if $I=\prod_{i=1}^r\mathfrak{P}_i^{e_i}$ where if $\{\mathfrak{p}_j\}=\left\{\mathfrak{P}_i\cap R\right\}_{i=1}^r$ has that

$$\mathfrak{p}_jS=\left(\prod_{k=1}^{e_j}\mathfrak{P}_{i_k}^{e'_{i_k}}\right)^n$$

for every $s$ where $ne'_{i_k}=e_{i_k}$ for all $k$.