Your diagrams that you have in the first order are all equivalent. However you shouldn't lose sight of the fact that Feynman diagrams are about pairing fields in Wick's theorem. There are 6 ways to pair the four $x$ fields in $x^4$ so this Feynman diagram will carry a factor of 6. We don't actually draw 6 diagrams but this combinatorial factor is still important.

In your second order diagrams, the third one is a disconnected diagram. If you expand ${Z[\lambda]}/{Z[0]}$ to 2nd order you indeed do get this disconnected diagram since it is a valid way of pairing the fields. However, you are interested in calculating $W$, $$\frac{Z[\lambda]}{Z[0]}=\exp[W]=\exp\left[\lambda W_1+\lambda^2 W_2+\dots\right]=1+\lambda W_1 +\lambda^2(\frac{1}{2!}W_1^2 + W_2)+\dots$$ The disconnected diagram is already accounted for by the $\lambda^2W_1^2/2$ term and so does not contribute to the new $W_2$ term.

This will be true in general for $W$. You can just ignore any disconnected diagram.

Your diagrams that you have in the first order are all equivalent. However you shouldn't lose sight of the fact that Feynman diagrams are about pairing fields in Wick's theorem. There are 3 ways to pair the four $x$ fields in $x^4$ so this Feynman diagram will carry a factor of 3. We don't actually draw 3 diagrams but this combinatorial factor is still important.

In your second order diagrams, the third one is a disconnected diagram. If you expand ${Z[\lambda]}/{Z[0]}$ to 2nd order you indeed do get this disconnected diagram since it is a valid way of pairing the fields. However, you are interested in calculating $W$, $$\frac{Z[\lambda]}{Z[0]}=\exp[W]=\exp\left[\lambda W_1+\lambda^2 W_2+\dots\right]=1+\lambda W_1 +\lambda^2(\frac{1}{2!}W_1^2 + W_2)+\dots$$ The disconnected diagram is already accounted for by the $\lambda^2W_1^2/2$ term and so does not contribute to the new $W_2$ term.

This will be true in general for $W$. You can just ignore any disconnected diagram.