The current of M5 and M7 of two stage OTA
This is one of those weird questions you see once and then, the very first words you say are "Let direct current = 100μ", so I am assuming that \$I_{DC}=100µA\$. Which flows straight down through \$M_8\$. And then you are asking about what you said, what the current is through \$M_8\$. 

Well... If you put \$100µA\$ through \$M_8\$ then... I suppose \$100µA\$ will flow through \$M_8\$. 

Remember, those are MOSFET's, not BJT's



EDIT

Woaw, I missed the questions under the image, I didn't see them. It's a messy question.

You say early in your question that you know that \$M_3=M_4=M_6=M_1=M_2=100μ\$

The \$M_4 = M_3\$ part is not 100% true. It's rather \$M_4 ≤ M_3\$. It will try to make a current mirror, but if it can't... then it will be less. 

You got two cases here. A is when \$V_{IN} &gt; V_{IP}\$ and B is when \$V_{IN} ≤ V_{IP}\$.



In case A, \$V_{IN} &gt; V_{IP}\$, this happens:

\$100µA = M_8 = M_5 = M_1 = M_3\$

\$≈0A = M_6 = M_7\$ and \$M_4 = M_2\$ and \$M_2 = 100µA - M_1 → M_2  ≈0A \$





In case B, \$V_{IN} ≤ V_{IP}\$, this happens:

\$100µA = M_8 = M_5 = M_7 = M_6\$

\$50µA = M_1 = M_3 = M_2 = M_4\$





Here is the simulation that you should've made. 
I've marked two voltages "Scroll", put your mouse over the voltages and scroll with your mouse to change the voltage and you will see what happens.