A large nonconducting sheet M carries a uniform charge density. Two...

Let the large nonconducting sheet $M$ have a positive charge density (the reasoning remains the same for a negative charge).

The electric field produced by a large (but finite) sheet decreases with distance.

This electric field polarizes the uncharged metal rods $A$ and $B$. For rod $A$, the end closer to $M$ acquires a negative charge, and the farther end acquires a positive charge.

Since the electric field of $M$ is stronger at the closer end, the attractive force on the negative charge is greater than the repulsive force on the positive charge. Thus, the net force is attractive, and $M$ attracts rod $A$.

Similarly, rod $B$ is polarized with a negative charge on its closer end and a positive charge on its farther end. Because the field of $M$ continues to decrease with distance, the attractive force on the closer end of $B$ is greater than the repulsive force on its farther end. Thus, $M$ also attracts rod $B$.

Furthermore, the right end of rod $A$ (positively charged) and the left end of rod $B$ (negatively charged) are adjacent to each other. Since they have opposite charges, they attract each other. Thus, rod $A$ attracts rod $B$.

Therefore, options (1), (3), and (4) are correct.

Answer: The sheet $M$ attracts rod $A$.; The sheet $M$ attracts rod $B$.; Rod $A$ attracts rod $B$.