Does $ Ev(\cos(4\pi t)u(t)) $ signal has period? According to Oppenheim's signal and systems solution book and other solutions like this

$$
Ev(\cos(4\pi t)u(t)) = 1/2 \times ( \cos(4\pi t)u(t) + \cos(4\pi t)u(-t)) = 1/2(\cos(4\pi t)) 
$$

$$
\textrm{for }-\infty &lt; t &lt;  \infty
$$

is periodic and its period $1/2$.

But I think it is not equal $ 1/2(\cos 4\pi t)  $ at $ t = 0 $ so it is not periodic because at $ t = 0 $ it is equal to $ 1 $ different from other multiples of period.

I have also curious about the value of $ u(t)+u(-t) $. Is it $0$, $1$, or $2$ ?
It is a matter of convention. The answer depends on the definition of the step function $u(t)$. In signal processing, it is common to define $u(0)=1/2$. Under this definition, we have that $Ev(\cos(4\pi t)u(t)) = 1/2(0.5+0.5)=1/2$, so the signal is indeed periodic.

If $u(0)$ is defined as either 0 or 1, then you'd be correct and the signal wouldn't be periodic.