$P(1)$ is true. If $P(k)$ is true then consider $P(k+2)$:

${\displaystyle 9^{k+2}+1^{k+2}=81(9^k+1^k)-80}$

so $P(k+2)$ is true and the result is true for all odd numbers by the axiom of induction.

Similarly $7^1+3^1$ is a multiple of 10 and if $7^k+3^k$ is a multiple of 10 then

${\displaystyle 7^{k+2}+3^{k+2}=49(7^k)+9(3^k)=40(7^k)+9(7^k+3^k)}$

so this is also a multiple of 10 and the result is true for all odd numbers by the axiom of induction. Note that $9^2+1^2$ and $7^2+3^2$ are not multiples of 10 so the result is not true for even numbers.

(2) We have $8^0-2^0$ and $8^2-2^2$ are both multiples of 10. If $8^k-2^k$ is a multiple of 10 then

${\displaystyle 8^{k+2}-2^{k+2}=64(8^k)-4(2^k)=60(8^k)+4(8^k-2^k)}$

so this is a multiple of 10 and the result is true for all even numbers. Similarly $6^2-4^2$ is a multiple of 10 and if $6^k-4^k$ is a multiple of 10 then

${\displaystyle 6^{k+2}-4^{k+2}=36(6^k)-16(4^k)=20(6^k)+16(6^k-4^k)}$

which is a multiple of 10 and so the result is true for all even numbers.

Note that $8-2$ and $6-4$ are not multiples of 10 and the result is not true for odd powers.

(3) The corresponding results are $9^n-1^n$ and $7^n-3^n$ are multiples of 10 when $n$ is an even number which can be proved by induction.

(4) Similarly $8^n+2^n$ and $6^n+4^n$ are multiples of 10 when $n$ is an odd number which can be proved by induction.

Proof using the Binomial Theorem

${\displaystyle 7^n+3^n=7^n+(10-7{)}^n=7^n+{10}^n+.....+(-7{)}^n.}$

All the terms of the Binomial expansion involve powers of 10 apart from the term $(-7{)}^n$ so the expression is a multiple of 10 when $n$ is odd but not when $n$ is even. The same method works for all other parts of the question.