Convert $31213111332_{4}$ to hexadecimal I did
$$31213111332_{4} = \\
2*4^0+3*4^1+3*4^2+1*4^3+1*4^4+1*4^5+3*4^6+1*4^7+2*4^8+1*4^9+3*4^{10} = \\
2*16^0+3*4^1+3*16^1+1*4*16^1+1*16^2+1*4*16^2+3*16^3+1*4*16^3+2*16^4+1*4*16^4+3*16^5 = \\
2*16^0 + 3*4^1+7*16^1+5*16^2+7*16^3+6*16^4+3*16^5 = \\
36757E_{16}$$
Which is correct but I have a question. What if $2*16^0+3*4^1$ was greater than 15 (F) ?

Don’t worry.  You would have at most $3*4^0+3*4^1=15\;(F)$, since base $4$ digits are at most $3.$ 
