Ideals in a ring satisfying $I \cap (J+K) \neq I \cap J + I \cap K$ I am not getting an example where $I \cap (J+K) \neq I \cap J + I \cap K$, where $I,J,K$ are ideals in a ring $A$.
 A: Take $A = k[x, y]$ where $k$ is a field, and take $I = (x)$, $J = (y)$, and $K = (y-x^2)$. Then $x^2$ belongs to $(x) \cap (y, y-x^2) = I \cap (J + K)$. On the other hand, $I \cap J$ is $(xy)$ (easy) and $I \cap K = (xy - x^3)$ (similarly easy by invoking an automorphism on $k[x, y]$ that takes $x$ to $x$ and $y$ to $y-x^2$), and one cannot write $x^2$ in the form $xyp(x, y) + (xy-x^3)q(x, y)$, so $x^2$ does not belong to $I \cap J + I \cap K$. 
The equality in the case of $\mathbb{Z}$ is easier to see by invoking the fundamental theorem of arithmetic, in the form that the lattice of ideals (ordered by reverse inclusion) is isomorphic to the set of functions $P \to \mathbb{N}$ with finite support (together with an element "top" that we adjoin as a top element of this lattice) where $P$ is the set of nonzero prime ideals of $\mathbb{Z}$; these functions are ordered by $f \leq g$ iff $f(p) \leq g(p)$ for every $p \in P$. Here the meet in the lattice of ideals is given by $\cap$ and the join by $+$; on the other side we just use $(f \wedge g)(p) := \min\{f(p), g(p)\}$ and $(f \vee g)(p) := \max\{f(p), g(p)\}$. Then the distributive law can be checked pointwise in $p$, and boils down to the easy claim that $\mathbb{N}$ under $\min$ and $\max$ satisfies the distributive law. 
