>Therefore, if m ≡ n mod 4, then 7\^m ≡ 7\^n mod 10.

This statement is just reiterating the previous statement, but more formally and using the modulus. You can essentially factor out 7\^4 of any higher power of 7 when working mod 10 since 7\^4 = 1 (mod 10), which is the same as saying that a difference of 4 in the exponent doesn't matter, which is the same as saying that the exponents can just be equal (mod 4).

An example: 7\^41 = 7\^49 (mod 10) because you can factor (7\^4)\^2 out of the right hand size, which we know to be 1 (mod 10).

>What is 7\^7 mod 4? Well 7\^1 ≡ 3 mod 4, 7\^2 ≡ 1 mod 4, so 7\^7 ≡ (7\^2)\^3 · 7 ≡ 3 mod 4.

This statement is just reducing 7\^7 mod 4. It does so by first noticing that 7 = 3 (mod 4) so 7\^7 = 3\^7. Then 3\^2 is calculated, then the cube of that which is 3\^6 = 7\^6 (mod 4), and then it's multiplied by 7 to get 7\^7 (mod 4). There's many other ways you could do it.

>7\^7\^7 ≡ 7\^3 ≡ 3 mod 10.

Here the exponent is bracketed as 7\^(7\^7). Using the rule established earlier, the exponent can be reduced mod 4 as the whole number is being reduced mod 10. So since 7\^7 = 3 (mod 4) that means that 7\^(7\^7) = 7\^3 = 3 (mod 10).