Why are these languages not regular?
I think it is easier to prove if you first note note $$ L' = \{a^nb^mc^k|n\leq k \land n,m,k&gt;0\} = \{a^nb^mc^k | n,m,k \geq 1 \} - \{a^nb^mc^k|n&gt;k \land n,m,k&gt;0\}$$ So if $\{a^nb^mc^k|n&gt;k \land n,m,k&gt;0 \}$ is regular then so is $\{a^nb^mc^k|n\leq k \land n,m,k&gt;0\}$ since $\{a^nb^mc^k | n,m,k \geq 1 \}$ is regular and the difference of regular sets is a regular set. 

But $L'$ is not regular, a contradiction. If you are NOT convinced that $L'$ is not regular then see the following proof using the Pumping lemma.


So, let's prove that $L' = \{a^nb^mc^k|n\leq k \land n,m,k&gt;0\}$ is not regular using the Pumping lemma in order to obtain contradiction.

The Pumping Lemma:

Let $L$ be a regular set. Then there is a constant $n$ such that if $z$ is any word in $L$, and $|z| \geq n$ we may write $z = uvw$ in such a way that $|uv| \leq n$, $|v| \geq 1$, and for all $i \geq 0$, $uv^iw \in L$. 

How to prove

1) Select the language you wish to to prove irregular. 

2) The "adversary" picks $n$, the constant mentioned in the Pumping lemma. Once it has been picked, it may not be changed (fix it).

3) Select a string $z$  in $L$. Your choice may depend implicitly on the value of $n$.

We choose $z = a^nb^nc^{n+1}$ (since $n &lt; n+1$, $z \in L$).

4) The "adversary" breaks $z$ into $u$, $v$, and $w$, subject to the constraints that $|uv| \leq n$ and $|v|$.

5) You achieve a contradiction to the pumping lemma by showing that there exists $i$ such that $uv^iw \notin L$.

Since $z = a^nb^nc^{n+1} = uvw$ and $|uv| \leq n$, $uv$ is a string consisting of only $a$s, and so is $v$. If we choose $i=n+2$ then we get 
$uvvw$ such that $uv^{n+2}w$ has at least $n+2$ $a$s which is greater than $n+1$ $c$s which does not belong to $L'$. But this contradicts to the Pumping lemma.

You may try to prove the second language  yourself.