Combining 2x1 mux and 4x1 mux with AND gate
Your answer is incorrect, since when computing the output of both multiplexers, you incorrectly took into the account the dependency of the output on the selector / selectors. Watch closely how I did it and compare to the results you got!
According to the truth table of a 2x1 mux and a 4x1 mux, we can write the following, and use boolean algebra rules in order to simplify:
The output of the 2x1 mux: $$Z_1=A' B+A\cdot 0=A'B$$
The output of the 4x2 mux: $$Z_2=A'B'\cdot C+A'B\cdot 0+AB' \cdot 1+AB \cdot 1 = A'B'C+AB'+AB=A'B'C+A$$
Then, it's clear that: $$F=Z_1Z_2=A'B(A'B'C+A)=A'BA'B'C+A'BA=0+0=0$$
Or, in other words, the output F is False.
P.S - I used this helpful site to verify the answer.