Answer
$a-(4,0)\\
b-(0,1)\\
c-(2,4)\\
d-(3,3)\\$
Work Step by Step
$G: J_{5} \times J_{5}\rightarrow J_{5} \times J_{5}\\
\,For\,all\,ordered\,pairs\,(a, b)\,\in J_{5}\times J_{5} \\
G(a, b) = ((2a + 1)\,mod\,5, (3b-2)\,mod\,5) \\
a-\\
G(4, 4) = ((2(4) + 1)\,mod\,5, (3(4)-2)\,mod\,5)\\=(9\,mod\,5,10\,mod\,5)=(4,0) \\
b-\\
G(2, 1) = ((2(2) + 1)\,mod\,5, (3(1)-2)\,mod\,5)\\=(5\,mod\,5,1\,mod\,5)=(0,1) \\
c-\\
G(3, 2) = ((2(3) + 1)\,mod\,5, (3(2)-2)\,mod\,5)\\=(7\,mod\,5,4\,mod\,5)=(2,4) \\
d-\\
G(1, 5) = ((2(1) + 1)\,mod\,5, (3(5)-2)\,mod\,5)\\=(3\,mod\,5,13\,mod\,5)=(3,3) \\$