Maximal chains have the same length? 1) If one has a finitely generated algebra over a field, is it true that any two maximal chains of primes have equal length?
2) If not, are there any other conditions that allows one to conclude that the maximal chains of primes have equal length?
 A: No, it is not true: a counterexample is $k[x,y,z]/(xz,yz)$.
It is true however for a finitely-generated algebra $A$ (over a field $k$) without zero-divisors .
Indeed this follows from [Matsumura, Commutative Rings, Ch.5, (14.H)].  
There he proves that $A$ is "catenary" (even "universally catenary", a stronger property) .
He also proves a formula which implies  that all maximal ideals of $A$ have height $dim(A)$, and together these results show  that all maximal chains of prime ideals have the same length, namely $dim(A)$.
[He defines "catenary" page 84] 
A: Note that the assumption over a field is very important here. One can prove that in a finitely generated domain over $k$ every maximal chain of primes have the same length using transcendence degree or Noether normalization.
Here is an interesting example to show that even if we consider regular rings finitely generated over a DVR, the statement is false. I strongly recommend you to remember the following example in mind as a sanity check.

Consider a DVR $R$ with uniformizer $a$, then $R[x]$ has the following two maximal chains of primes $0 \subset (a) \subset (a,x)$ and $0 \subset (ax-1)$. Note that $(ax-1)$ is maximal because it inverts $a$, and thus we get $K(R)$ as the quotient.

A: One of the important condition is Cohen-Macaulay ring. 
see https://stacks.math.columbia.edu/tag/00N7, Lemma 10.103.3. 
The first statement is not true, even though it is catenary. The easy example is in $\mathbb{R}^3$, consider the union of a hyperplane and a line intersecting. It is also a standard counterexample of non-MC rings. 
