jmc

$2^{12}\times 5^8=2^4\times (2\times 5{)}^8=16\times 10^8$. $10^8$ has 9 digits, so $16\times (10{)}^8$ has 10 digits (1, 6, and eight 0's). Therefore, there are $\boxed{10}$ digits in the value of $2^{12}\times 5^8$.