You can have $0 ... 3$ $C$'s.
One way to arrange three $C$'s: $CCC$.
Three ways to arrange two $C$'s and another consonant: $CCX, CXC, XCC$. $X$ is one of $R,F,M,N$, so $12$ total for this case.
Six ways to arrange one $C$ with two different consonants: $CXY, CYX, XYC, YXC, XCY, YCX.$ There are $_4C_2 = 6$ ways to pick two of four consonants, so $36$ total ways for this case.
Three ways to arrange one $C$ with the repeated consonant: $CRR, RCR, RRC$.
If there are no $C$, you can have two $R$'s or not.
There are nine ways to arrange two $R$'s and one non-$C$ consonant.
There are $24$ ways to arrange three different non-$C$ consonants.
So ... adding everything gives $1+12+36+3+9+24 = 85$ ways to arrange three consonants.
(I think you can actually get $13$ arrangements of vowels: $eee, eei, eie, iee, eeu, eue, uee, ieu, iue, eiu, eui, uie, uei$.)
To continue then, pick the arrangement of consonants $(85)$, pick the arrangement of vowels $(13)$ and then pick the three places where the consonants will go in the order you've chosen $(_6C_3 = 20)$. (The vowels fall into the remaining spaces in the order you've chosen already.)
The total number of $6$-letter words then is $85 \cdot 13 \cdot 20 = 22100.$
For the second part of the question, there are $4$ ways to have three $C$'s together, and $13$ orderings for the vowels, so $52$ total ways.