Problem 1 is fully factored as each term is a binomial raised to some exponent. If the exponent isn't showing up, it's because it is 1. Recall that x^1 = x.
Problem 2 can be factored further because x^2-8x+16 factors to (x-4)(x-4) or (x-4)^2. To get this factorization, you find two numbers that multiply to 16 and add to -8. Those two numbers are -4 and -4 which is where the (x-4)(x-4) comes from. Overall, the entire thing factors to (x-4)^2*(x+3)*(x-2)
Problem 3 is a similar story. We can factor x^2-1 into (x-1)(x+1). I used the difference of squares rule here. Or you can think of x^2-1 as x^2+0x-1, then find two numbers that multiply to -1 and add to 0. Those two numbers are +1 and -1 which leads to (x+1)(x-1). So the full factorization is (x-1)(x+1)(x+1)(x-4) which is the same as (x-1)(x+1)^2(x-4)
