Neither 2 nor 7 have inverses mod 6 or 7, respectively, so the expressions in (a) and (b) cannot really evaluated... At least we can evaluate (c) :
[tex]5\times3 \equiv 15 \equiv 1 \pmod 7 \\\\ \implies 3\times2\div5 \equiv 3\times2\times3 \equiv18 \equiv \boxed{4 \pmod{7}}[/tex]
