First-order logic - Negations

Question

I have the following:

The language $L_{geom}$ of basic plane geometry has two 1-place relation symbols $P$ and $L$ for “point” and “line”, and a 2-place relation symbol $I$ for “point $x$ lies on line $y$”.

1. $\forall x\Big(\big(P(x)\lor L(x)\big)\land \lnot\big(P(x)\land L(x)\big)\Big)$
2. $\forall x\forall y\Big(\big(P(x)\land P(y)\big)\to\exists z\big(L(z)\land I(x,z)\land I(y,z)\big)\Big)$
3. $\forall x\forall y\Big(\big(P(x)\land P(y)\big)\to\exists z\big(L(z)\land I(x,z)\land I(y,z)\land\forall w(L(w)\land I(x,w)\land I(y,w)\to w=z)\big)\Big)$

and the negation (without the symbol $\lnot$ appearing in the result) I got in each case was:

1. In this case I don't know how to do it
2. $\exists x\exists y\Big(P(x)\land P(y)\to\forall z\big(I(x,z)\lor I(y,z)\big)\Big)$
3. $\exists x\exists y\Big(P(x)\land P(y)\to \forall z\big(L(z)\land I(x,z)\land I(y,z)\to\exists w(L(w)\land I(x,w)\land I(y,w)\land w\ne z)\big)\Big)$