Let S be the set of all real numbers for $a,b\in S,$

Statement–I: Relation R is defined by aRb iff $1+ab>0$ then R is reflexive and symmetric
Statement–II: Relation R is defined by aRb iff $|a-b|<1$ then R is an equivalence relation

The given statement is 

Statement–I: Relation R is defined by $aRb$ iff $1+ab>0$ then $R$ is reflexive and symmetric

Let $a\in R\Rightarrow 1+a.a=1+a^2>0$

R is reflexive

Let $(a,b)\in R\Rightarrow 1+ab>0\Rightarrow 1+ba>0$

R is symmetric

Since $(1,\frac{1}{3})\in R$ and $(\frac{1}{3},-1)\in R$ but $(1,-1)\notin R$

R is not transitive

Therefore, the statement I is true. 

Consider the statement II: 

Statement–II: Relation R is defined by aRb iff $|a-b|<1$ then R is an equivalence relation.

Let $a\in R\Rightarrow \left|a-a\right|=0<1$

The relation $R$ is reflexive.

Let $(a,b)\in R$

$aRb\Rightarrow |a-b|<1\Rightarrow |b-a|<1\Rightarrow bRa\Rightarrow R$ is symmetric

Suppose that $a=1,b=\frac{1}{2},c=-\frac{1}{4}$

$\left|1-\frac{1}{2}\right|<1,\Rightarrow \left(1,\frac{1}{2}\right)\in R$

$\left|\frac{1}{2}+\frac{1}{4}\right|<1,\Rightarrow \left(\frac{1}{2},-\frac{1}{4}\right)\in R$

But $\left(1,-\frac{1}{4}\right)\notin R$ because $\left|1+\frac{1}{4}\right|>1$

Hence, the relation is not transitive'

So that the relation R is not equivalence

The statement II is False

Hence, the statement -I is true, and Statement - II is False.