Alkanes or $\mathrm{sp^3}$ hybridized carbons show +I effect.
Alkenes contain $\mathrm{sp^2}$ hybridized carbons. Due to greater $\mathrm{s}$ character they have higher electron withdrawing tendency. So they show -I effect.
Coming to the first question:
In the given figures we have $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized carbons in the given rings. Notice that in the second figure there are greater number of $\mathrm{sp^2}$ hybridized carbons. Hence, the net withdrawal of electron density is more, and accordingly, stability goes down. In the third figure there is no resonance, which is the more dominating effect than the inductive effect. So it is the least stable.
In the second question:
In figures 1 and 2, we have equal number of $\mathrm{sp^2}$ hybridized carbons. But notice in the first figure, the negative charge is on the ring and is closer to the $\mathrm{sp^3}$ carbons which are electron donating (Remember, inductive effects are distance dependent). In the third figure greater +I is being imposed on the anion, hence it has the least stability.