Answer
$ J_{5} = \left \{ 0, 1, 2, 3, 4 \right \},
h :J_{5}\rightarrow J_{5}
\,\,and\,\, k :J_{5}\rightarrow J_{5}\\
h (x)=(x + 3)^{3}\,\,mod\,\,5\,\,and\,\,k(x)=(x^3+4x^{2} + 2x + 2)\,\,mod\,\,5.\\
h (0)=(0 + 3)^{3}\,\,mod\,\,5=27\,mod\,5=2\\
h (1)=(1 + 3)^{3}\,\,mod\,\,5=64\,mod\,5=4\\
h (2)=(2 + 3)^{3}\,\,mod\,\,5=125\,mod\,5=0\\
h (3)=(3 + 3)^{3}\,\,mod\,\,5=216\,mod\,5=1\\
h (4)=(4 + 3)^{3}\,\,mod\,\,5=343\,mod\,5=3\\
k(0)=(0^3+4(0)^{2} + 2(0) + 2)\,\,mod\,\,5=2mod\,5=2\\
k(1)=(1^3+4(1)^{2} + 2(1) + 2)\,\,mod\,\,5=9mod\,5=4\\
k(2)=(2^3+4(2)^{2} + 2(2) + 2)\,\,mod\,\,5=30mod\,5=0\\
k(3)=(3^3+4(3)^{2} + 2(3) + 2)\,\,mod\,\,5=71mod\,5=1\\
k(4)=(4^3+4(4)^{2} + 2(4) + 2)\,\,mod\,\,5=138mod\,5=3\\
so\,\,\\
h(0)=k(0)=2 \\
h(1)=k(1)=4 \\
h(2)=k(2)=0 \\
h(3)=k(3)=1 \\
h(4)=k(4)=3 \\
\therefore h=k
$
Work Step by Step
$ J_{5} = \left \{ 0, 1, 2, 3, 4 \right \},
h :J_{5}\rightarrow J_{5}
\,\,and\,\, k :J_{5}\rightarrow J_{5}\\
h (x)=(x + 3)^{3}\,\,mod\,\,5\,\,and\,\,k(x)=(x^3+4x^{2} + 2x + 2)\,\,mod\,\,5.\\
h (0)=(0 + 3)^{3}\,\,mod\,\,5=27\,mod\,5=2\\
h (1)=(1 + 3)^{3}\,\,mod\,\,5=64\,mod\,5=4\\
h (2)=(2 + 3)^{3}\,\,mod\,\,5=125\,mod\,5=0\\
h (3)=(3 + 3)^{3}\,\,mod\,\,5=216\,mod\,5=1\\
h (4)=(4 + 3)^{3}\,\,mod\,\,5=343\,mod\,5=3\\
k(0)=(0^3+4(0)^{2} + 2(0) + 2)\,\,mod\,\,5=2mod\,5=2\\
k(1)=(1^3+4(1)^{2} + 2(1) + 2)\,\,mod\,\,5=9mod\,5=4\\
k(2)=(2^3+4(2)^{2} + 2(2) + 2)\,\,mod\,\,5=30mod\,5=0\\
k(3)=(3^3+4(3)^{2} + 2(3) + 2)\,\,mod\,\,5=71mod\,5=1\\
k(4)=(4^3+4(4)^{2} + 2(4) + 2)\,\,mod\,\,5=138mod\,5=3\\
so\,\,\\
h(0)=k(0)=2 \\
h(1)=k(1)=4 \\
h(2)=k(2)=0 \\
h(3)=k(3)=1 \\
h(4)=k(4)=3 \\
\therefore h=k
$