I am wonder whether my answer of these questions is correct compared with the original answer.
Let L(x,y) be the statement"x loves y," where the domain for both x and y consist of all people in the world.
1.There is exactly one person whom everybody loves.
Original answer:∃x(∀yL(y,x)∧∀z(∀wL(w,z)⟹z=x))
My answer:∃x∀y∃z(L(y,x)∧((x≠y)⟹⌐L(z,y)))
My thought is that there exists a x that every y love,if y is not x,there must have a z doesn't love y,so there is exactly one person whom everybody loves
2.There is exactly two people whom Lynn loves.
My answer:∃x∃y∀z(L(Lynn,x)∧L(Lynn,y)∧(x≠y)∧(((x≠z)∧(y≠z))⟹⌐L(Lynn,z)))
My thought is that there exists a x and y(they are not the same) that Lynn love,for every z that is not equal to x and y,then Lynn doesn't love,so there is exactly two people whom Lynn loves.
Thank for your generous.
Your answer 1 is actually equivalent to the original answer: \begin{aligned} &\exists x\forall y\exists z(L(y,x)\land ((x\ne y)\implies\lnot L(z,y))) && \text{your formula}\\ \iff & \exists x\forall y (L(y,x)\land \exists z((x\ne y)\implies\lnot L(z,y))) && \text{$L(y,x)$ does not contain $z$}\\ \iff & \exists x\forall y (L(y,x)\land \exists z(L(z,y)\implies x=y)) && \text{reversing implication}\\ \iff & \exists x (\forall y L(y,x)\land \forall y\exists z(L(z,y)\implies x=y)) && \text{$\forall$ distributes over $\land$}\\ \iff & \exists x (\forall y L(y,x)\land \forall y\exists w(L(w,y)\implies x=y)) && \text{variable renaming}\\ \iff & \exists x (\forall y L(y,x)\land \forall z\exists w(L(w,z)\implies x=z)) && \text{variable renaming}\\ \iff & \exists x (\forall y L(y,x)\land \forall z\exists w(L(w,z)\implies z=x)) && \text{symmetry of equality}\\ \iff & \exists x (\forall y L(y,x)\land \forall z\exists w(\lnot L(w,z)\lor z=x)) && \text{rewrite implication}\\ \iff & \exists x (\forall y L(y,x)\land \forall z(\exists w\,\lnot L(w,z)\lor z=x)) && \text{$z=x$ does not contain $w$}\\ \iff & \exists x (\forall y L(y,x)\land \forall z(\lnot\forall w L(w,z)\lor z=x)) && \text{move quantifier through $\lnot$}\\ \iff & \exists x (\forall y L(y,x)\land \forall z(\forall w L(w,z)\implies z=x)) && \text{rewrite to implication} \end{aligned} The last line is exactly the original answer.
Your answer 2 is also correct.
I notice that in both cases, you put all the quantifiers to the beginning. From a logical standpoint, there's nothing wrong with this, however generally formulas are easier to understand if the scope of quantifiers is as narrow as possible.
For this reason, I would advise you to narrow the scope of each variable as much as possible. Note that the original answer for 1 does this.
In your answer 2, only the last clause of the conjunction uses $z$, therefore you can put that quantifier there: $$\exists x\exists y(L(\text{Lynn},x)\land L(\text{Lynn},y)\land\forall z(((x\ne z)\land(y\ne z))\implies\lnot L(\text{Lynn},z)))$$