Solution: 1) Proof: from: f (x+2) f (x) =1:f (x) =1/f (x+2)

Assign the x value of the above formula to x+2, that is, f(x+2)= 1/f(x+4).

So: f (x) =1/f (x+2) =1[1/f (x+4)] = f (x+4).

Therefore, the function f(x) is a periodic function with a period of 4.

2) The function f(x) is a periodic function with a period of 4, 1 19/4=29+3.

So f( 1 19)=f(3).

Where: f(x+2)f(x)= 1, x =1:f (3) f (1) =1.

Let x =-1:f (1) f (-1) =1.

Since f(x) is an even function defined on r, f( 1)=f(- 1).

So, f? (1)= 1, and f(x) is greater than 0.

Therefore, f( 1)= 1.

So f (119) = f (3) =1/f (1) =1.

I hope my answer is helpful to you. Good luck.