Is the group $U(8)$ cyclic?

I guess you are mixing $\mathbb{Z_8}$ and $U(8)$. $1$ is identity in $U(8)$ as operation is multiplication here. But in $\mathbb{Z_8}$ operation is addition, so $0$ is the identity and $1$ is the generator in $\mathbb{Z_8}$ as you can have any elements less than $8$ by adding $1$. No $U(8)$ is not cyclic. To see it pick an element of $U(8)$ and keep multiplying it with itself. If you find one with order 4, that is your generator. but every element here is of order 2.

$\textbf{Theorem-}$ $U(n)$ is cyclic iff $n$= $2,4, p^k,2p^k $ where $p$ is an odd prime. Try to prove it.