Solution:

a. Let $a, b, c$ be the sides of a triangle with $a + b + c = 1996$, and each being a positive integer. Then $a + 1, b + 1, c + 1$ are also sides of a triangle with perimeter $1999$ because
$$a < b + c \quad \Longrightarrow \quad a + 1 < (b + 1) + (c + 1)$$
and so on. Moreover, $(999, 999, 1)$ form the sides of a triangle with perimeter $1999$, which is not obtainable in the form $(a + 1, b + 1, c + 1)$ where $a, b, c$ are the integers and the sides of a triangle with $a + b + c = 1996$. We conclude that $f(1999) > f(1996)$.

b. As in the case (a) we conclude that $f(2000) \geq f(1997)$. On the other hand, if $x, y, z$ are the integer sides of a triangle with $x + y + z = 2000$, and say $x \geq y \geq z \geq 1$, then we cannot have $z = 1$; for otherwise we would get $x + y = 1999$ forcing $x, y$ to have opposite parity so that $x - y \geq 1 = z$ violating triangle inequality for $x, y, z$. Hence $x \geq y \geq z > 1$. This implies that $x - 1 \geq y - 1 \geq z - 1 > 0$. We already have $x < y + z$. If $x \geq y + z - 1$, then we see that $y + z - 1 \leq x < y + z$, showing that $y + z - 1 = x$. Hence we obtain $2000 = x + y + z = 2x + 1$ which is impossible. We conclude that $x < y + z - 1$. This shows that $x - 1 < (y - 1) + (z - 1)$ and hence $x - 1, y - 1, z - 1$ are the sides of a triangle with perimeter $1997$. This gives $f(2000) \leq f(1997)$. Thus we obtain the desired result.