Part 1:
x/y > 1 by definition, and this is true iff x>y (since we are working over the natural numbers) which is itself true iff x+1 > y+1, which is true iff (x+1)/(y+1) > 1.
Therefore, x/y > 1 implies (x+1)/(y+1)>1.
It remains to show that x/y > (x+1)/(y+1), which is true iff x(y+1) > (x+1)y, which is true iff xy+x > xy+y, which is true iff x>y.
Therefore, x/y > 1 implies x/y > (x+1)/(y+1) > 1.
Considering P, we see that 2/1 > 3/2 as shown above, so if we replace every second element k/(k-1) of the product with (k+1)/k, we get:
P^2 = 2/1*2/1 * 4/3*4/3 * 6/5*6/5... *k/(k-1)
P^2 > 2/1*3/2 * 4/3*5/4 * 6/5*7/6... *k/(k-1)*(k+1)/k= 3/1*5/3*7/5...*(k+1)/(k-1) == k+1
so P^2 > k+1 and so P > sqrt(k+1).
Part 2: From the result in part 1, we see that Q > sqrt(101), but this is too weak a bound.
We can improve the bound by removing some of the worst approximations: specifically, that 2/1 > 3/2 (while this is true, it is very rough). Taking Q * 1/2 * 3/4 * 5/6 = 8/7*10/9...*100/99, so that (5/16 Q)^2 > 101/7 as before, we find that Q^2 > 101/7 * 256/25 > 101/7 * 10 = 1010/7 > 144, so q > 12.