a.$$\forall x\forall y \exists z (x\neq y)\rightarrow (x\neq z)$$ b. $$\neg\exists x\forall y \forall z (x=y)\rightarrow (x=z)$$
To revoke a. we need to find a case of $(x\neq z)\land (x=y)$ and for some $x,y$ (x=y) is false, therefore $(x\neq y)\land (x=z)$ is false and it always true.
b. is $\forall x\exists y \exists z (x\neq z)\lor (x=y)$ and it is always true, because for all $x$ there $y$ such $x=y$
On
One has to be careful with the possible domains for formulae like this. For (a): If the domain has two or more distinct elements x, y, then as already posted one can take z=y. If the domain has only one element, then (a) is still true, but now it is true vacuously because the condition (x $\neq$ y) can never be satisfied. If the domain is empty, however, then $\exists$z is always false regardless of the expression that follows. Conclusion: NOT A TAUTOLOGY. [I leave (b) for the student]
a) This seems a tautology. Just take $z=y$.
b) I think you have mistaken in negating the implication.