Why can't I use $I_1\omega_1=I_2\omega_2$ in this case? Since $I$ was taken of the entire system in the first problem. So why can't I just take $I$ of the system in this case? Which would obviously give the equation $0=\left(\frac{MR^2}{2}+mR^2\right)\omega$ where $m$ is the mass of the person.

So what are you going to plug in for $\omega$? The man and the disc have different angular velocities. If we consider the man is moving clockwise, the disc is rotating counter-clockwise. Therefore they don't have a common $\omega$.

But in your first question, all bodies have the same angular velocity.

Note that the equation $I_1\omega_1=I_2\omega_2$ is one form of the the original equation which is $L_{\text{initial}}=L_{\text{final}}$.

Why can't I use $I_1\omega_1=I_2\omega_2$ in this case? Since $I$ was taken of the entire system in the first problem. So why can't I just take $I$ of the system in this case? Which would obviously give the equation $0=\left(\frac{MR^2}{2}+mR^2\right)\omega$ where $m$ is the mass of the person.

So what are you going to plug in for $\omega$? The man and the disc have different angular velocities. If we consider the man is moving clockwise, the disc is rotating counter-clockwise. Therefore they don't have a common $\omega$.

But in your first question, all bodies have the same angular velocity.

Actually your second equation is not incorrect. It will give you $\omega=0$ which is the angular momentum of the system, as expected. Though you are asked to find the angular velocity of the disc. Therefore you have to consider the components of the system (man and disc) seperatedly.