It is false that any nilpotent element of a ring is contained in every ideal. For example, in the ring $R[x,y]/(x^n)$ where $R$ is any ring and $n>1$, the element $x^n=0$ is every ideal but $x^r$ is not in any ideal of the form $(g)$ for $g\in R[y]$ when $r<n$, for example. To be even more concrete, consider the example of $\mathbb{Z}[x]/(x^2)$, in which $x^2=0$ but $x$ is not in the ideal generated by $2$, for example.
Also, "smallest non-zero ideal of the ring" doesn't make sense for most rings. For example, in $\mathbb{Z}$, there is no minimal non-zero ideal; given any non-zero ideal $(n)$, the ideal $(2n)$ is smaller, and still non-zero. So there are no minimal non-zero ideals in $\mathbb{Z}$. In general, the poset of non-zero ideals may have many minimal elements: if $n\in\mathbb{Z}$ is a product of $k$ distinct prime numbers $q_1,\ldots,q_k$, then $\mathbb{Z}/n\mathbb{Z}$ has $k$ distinct minimal non-zero ideals, the $a_i\mathbb{Z}/n\mathbb{Z}$ where $a_i=q_1q_2\cdots q_{i-1}q_{i+1}\cdots q_k$.
What is true is that the nilradical of $R$ is the intersection of all the prime ideals of the ring $R$ (and this includes the zero ideal if $R$ is an integral domain).
This is because, if $P\subset R$ is a prime ideal, then $x^n=0\in P$ does imply that $x\in P$ (by the definition of prime ideal), so $$x\in\text{nil}(R)\implies x\in\bigcap_{P\subset R \text{ prime} }P,$$
and if $x\notin\text{nil}(R)$, then the collection $\Sigma$ of ideals of $R$ not containing $1,x,x^2,\ldots$ is a partially ordered set under inclusion, and it has some maximal element $M$ (using Zorn's lemma). This maximal element $M$ must be a prime ideal, because if $a\notin M$ and $b\notin M$, then $M+(a)$ and $M+(b)$ are both ideals of $R$ strictly containing $M$, hence containing powers of $x$ (because $M$ is maximal among ideals not containing powers of $x$). Thus the ideal $M+(ab)\supseteq (M+(a))(M+(b))$ contains a power of $x$, hence $M+(ab)$ strictly contains $M$, hence $ab\notin M$. Thus we have shown that $a,b\notin M\implies ab\notin M$, so $M$ is a prime ideal not containing $x$, and therefore we have shown
$$x\notin\text{nil}(R)\implies x\notin\bigcap_{P\subset R \text{ prime} }P.$$
Therefore
$$\text{nil}(R)=\bigcap_{P\subset R \text{ prime} }P.$$
Amitesh's exercises are also excellent, as usual :)
Certainly not. If we assumed that the intersections of infinitely many prime ideals always had to be prime, then the nilradical of any ring with infinitely many prime ideals would have to be prime, but the nilradical of a commutative ring is prime iff the ring has a unique minimal prime ideal. There are lots of counterexamples to that, though, so the original assumption was wrong.
Michalis pointed out that the intersection of two primes does not have to be prime. A simple example would be $F\times F$ for a field $F$. $F\times \{0\}$ and $\{0\}\times F$ are both maximal, hence prime, ideals, and their intersection is $\{0\}\times\{0\}$ which is certainly not prime.
This idea generalizes to an infinite ideal example. Let $R=\prod_{i=1}^\infty F$ For a field $F$. The ideal $M_i$ which is $F$ on every coordinate except on the $i$th coordinate, and is zero on the $i$th coordinate is a maximal, hence prime ideal of $R$. The intersection of these ideals is the zero ideal, but again, the zero ideal isn't prime.
Finally, let's get one with a nonzero nilradical. Take the same ring $R$ as above, and construct the triangular matrix ring of matrices that look like this: $\begin{bmatrix}a&b\\0&a\end{bmatrix}$ for $a,b\in R$. The intersection of maximal ideals of this ring is equal to the set of matrices of the form $\begin{bmatrix}0&b\\0&0\end{bmatrix}$ for $a,b\in R$, which is again not prime. There are infinitely many such matrices if your field is infinite.
However, there is an important case when the intersection of a family of prime ideals is prime: if they form a descending chain.
Best Answer
$\{0\}$ is not a prime ideal if $A$ has divisor of zeros. This also answer the second question, because if $I_{max}=\{0\}$, this means that $A$ is integral.