a) Let [tex]n=0[/tex]; then [tex]7^4-1=2400=30\cdot80[/tex], which is divisible by 30. We have
[tex]7^{n+4}-7^n=7^n(7^4-1)[/tex]
and since we know [tex]30\mid7^4-1[/tex], it follows that [tex]30\mid7^{n+4}-7^n[/tex] for all [tex]n\ge1[/tex].
b) We can write
[tex]3^{n+2}-2^{n+2}+3^n-2^n=(3^{n+2}+3^n)-(2^{n+2}+2^n)[/tex]
[tex]=3^n(3^2+1)-2^n(2^2+1)[/tex]
[tex]=10\cdot3^n+5\cdot2^n[/tex]
[tex]=10\cdot3^n+10\cdot2^{n-1}[/tex]
[tex]=10\left(3^n+2^{n-1}\right)[/tex]
and we're done.
