The way I thought about this question goes like this:
i) Look at the quotient $R/J_n = K[X,Y,Z]/(X^3, XY,Y^n)$. It looks like $K$ adjoin elements $a,b$ with $a^3 = ab = b^n = 0$. Now $1, a,a^2,b,b^2,...,b^{n-1}$ form a $K$-basis for this algebra. Given an element in $R/J_n$ , either it has its constant coefficient (that of the basis element $1$) nonzero, in which case it's not a zerodivisor; or it has only terms in the powers of $a$ and $b$, which means it will be nilpotent. So all zerodivisors in the quotient are indeed nilpotent.
ii) You were okay with this bit: intersection of two primary ideals, neither contained in the other, different radicals. Cool cool.
iii) Just use part i). Given any $n$, $I = (X^3,XY) = (X)\cap(X^3,Y) = (X)\cap(X^3,Y)\cap (X^3,XY,Y)\cap...\cap(X^3,XY,Y^n) $ just intersecting with as many $J_n$ as you fancy, because they all contain $(X^3,Y)$ so can't affect what the intersection actually is. This gives infinitely-many primary decompositions, but of course they're not minimal. Hope I understood that part of the question right!
As for finding the radical of $J_n$, I'm looking for the smallest prime ideal $P$ containing $J_n$. So, the quotient $R/P$ must be an integral domain, in which all elements of $J_n$ from $R$ are zero. So, $X^3 = Y^n = 0$ implies $\bar{X}$ and $\bar{Y}$ are both zerodivisors, and so need to be zero. So, $P$ contains $X$ and $Y$. But of course, $R/(X,Y)$ is an integral domain - isomorphic to $K$ - so $P = (X,Y)$ must be the radical.
Hope this was helpful, in spite of the messiness.
To flesh out the helpful comment by user26857, everything in the book is correct, it just requires a little work to see. The claim is that $(\mathfrak a^c:\mathfrak b^c)$ is a contraction, namely $(\mathfrak a:\mathfrak b^{ce})^c.$ This is rather easy to see, I guess.
\begin{align*}
(\mathfrak a^c:\mathfrak b^c) &= \{x\in B : xf^{-1}(\mathfrak b)\subseteq f^{-1}(\mathfrak a)\}\\
&=\{x\in B : xy\in f^{-1}(\mathfrak a) \text{ for all }y\in f^{-1}(\mathfrak b)\}\\
&=\{x\in B : f(xy)=f(x)f(y)\in \mathfrak a \text{ for all }y \text{ with } f(y)\in\mathfrak b\}\\
&=\{x\in B : f(x)z\in \mathfrak a \text{ for all }z\in f(f^{-1}(\mathfrak b))\}\\
&=\{x\in B : f(x)z\in \mathfrak a \text{ for all }z\in (f(f^{-1}(\mathfrak b)))\}\\
&=\{x\in B : f(x)z\in \mathfrak a \text{ for all }z\in \mathfrak b^{ce}\}\\
&=\{x\in B : f(x)\in (\mathfrak a:\mathfrak b^{ce})\}\\
&=(\mathfrak a:\mathfrak b^{ce})^c
\end{align*}
I guess is this a lesson on just how terse this book actually is.
Best Answer
In a Dedekind domain, for instance the ring of integers of a number field, then if $I$ and $J$ are ideals with $I\subseteq J$ then $I=J(I:J)$. In this case $(I:J)$ really is a quotient in the multiplicative sense, an ideal $X$ solving the equation $XJ=I$.