Finding if the given language is regular or not
$L$ is regular. 

Here is a proof as hinted by your second method, which points out that we may take advantage of the fact that regular languages are closed under complement. Another useful fact is regular languages is closed under intersection. Let $F=\{a^mb^nc^o| m,n,o\in\Bbb Z_{\ge 0}, m+n+o\le 5\}$. As a finite language, $F$ is regular. Let $H = \{a^mb^nc^o| m,n,o\in\Bbb Z_{\ge 0}\}$. As expressed by the regular expression $a*b*c*$, $H$ is regular. Since $L= H\cap\overline F$, $L$ is regular.

We can also see that $L$ is regular by the following one-line regular expression that expresses $L$. It is wrapped into multiple lines for easier reading.

$
((a\{6,\}|a\{5,\}b+|a\{4,\}b\{2,\}|a\{3,\}b\{3,\}|a\{2,\}b\{4,\}|a+b\{5,\})c*)|\\
((a\{5,\}|a\{4,\}b+|a\{3,\}b\{2,\}|a\{2,\}b\{3,\}|a+b\{4,\}|b\{5,\})c+)|\\
((a\{4,\}|a\{3,\}b+|a\{2,\}b\{2,\}|a+b\{3,\}|b\{4,\})c\{2,\})|\\
((a\{3,\}|a\{2,\}b+|a+b\{2,\}|b\{3,\})c\{3,\})|\\
((a\{2,\}|a+b+|b\{2,\})c\{4,\})|\\
(a+|b+)c\{5,\}|\\
c\{6,\}
$

Now that we have shown $L$ is a regular, the first method must be incorrect. In fact, we do not have to keep track of how many a′s, b′s and c′s we have seen. All we need is to make sure that the total number of them is greater than 5. There is a difference between these two requirements.



By the same argument, we can see easily that for any integer $k$, $L_k = \{a^mb^nc^o|m,n,o\in\Bbb Z_{\ge0},\,m + n + o &gt; k\}$ is a regular language.

Here is another similar fact. For any integer $s$ and $t$, $L_{s,t}=\{a^mb^nc^od^p|m,n,o,p\in\Bbb Z_{\ge0},\,m + o &gt; s, n+p &gt;t\}$  is a regular language.