$$c_1 = m_1 \oplus y_1 \text{ and } c_2 = m_2 \oplus y_2$$

will it be possible to obtain

$$z_1= (m_1 \lor m_2) \oplus y_1 \oplus y_2 \text{ and } z_2= (m_1 \land m_2) \oplus y_1 \oplus y_2$$

from $c_1$ and $c_2$ without revealing $m_1$, $m_2$, $y_1$ and $y_3$?

No, because then you could calculate $z_1 \oplus z_2 = (m_1 \land m_2) \oplus (m_1 \lor m_2) = m_1 \oplus m_2$.
In practice, you can only find $m_1 \oplus m_2$ if both $m_1$ and $m_2$ are encrypted with the same OTP (i.e., $(m_1 \oplus y_1) \oplus (m_2 \oplus y_1) = m_1 \oplus m_2$).
So without any knowledge of $m_1$, $m_2$, $y_1$ or $y_2$, there is no way of magically eliminating both of the OTP key streams $y_1$ and $y_2$.