Proof language A= { 0^i 1^j where j>(i mod 3), j ≥0 , i≥0} is regular language
$A$ is regular since it's the union of the following $3$ regular languages:


$\{0^i 1^j : i \bmod{3} = 0, j &gt; 0 \}$ for which a regular expression is $(000)^*11^*$.
$\{0^i 1^j : i \bmod{3} = 1, j &gt; 1 \}$ for which a regular expression is $(000)^*0111^*$.
$\{0^i 1^j : i \bmod{3} = 2, j &gt; 2 \}$ for which a regular expression is $(000)^*001111^*$.


A reguar expression for $A$ is then $E = (000)^*(1 + 011 + 00111)1^*$. You can now easily construct a NFA from $E$.