Solution:

a. $2005^{2004}$ is the sum of two positive perfect squares.
Observe that $5^{2} = 3^{2} + 4^{2}$ and that $2005^{2004} = 5^{2} m^{2}$, where $m = 5^{1001} \cdot 401^{1002}$. Multiplying the first relation by $m^{2}$ we obtain
$$2005^{2004} = (5m)^{2} = (3m)^{2} + (4m)^{2}.$$

b. $2004^{2005}$ is not the sum of two positive perfect squares.
First observe that $2004^{2005} = 3^{2005} \cdot 668^{2005}$ is divisible by $3$. Checking the remainder of the division by $3$ of the square of an integer $x$, we see that if $x$ is divisible by $3$ the remainder is $0$, while if it is not, and hence $x = 3k \pm 1$, then $x^{2} = 9k^{2} \pm 6k + 1$, so the remainder is $1$. If there existed positive integers $x, y$ such that $2004^{2005} = x^{2} + y^{2}$, then $x^{2} + y^{2}$ would be divisible by $3$, and hence so would the sum of the remainders of $x^{2}$ and $y^{2}$. Analyzing all the cases, we see that the only possibility is that both $x$ and $y$ are divisible by $3$. Now set $x = 3x_{1}, y = 3y_{1}$. Simplifying the equation $2004^{2005} = x^{2} + y^{2}$ by the common factor $9$, we obtain $3^{2003} \cdot 668^{2005} = x_{1}^{2} + y_{1}^{2}$. Analyzing divisibility by $3$ again, we find that $x_{1}$ and $y_{1}$ must also be divisible by $3$, and so the equation can again be simplified by dividing by $9$. Proceeding in this way, we will reach the point where the equation reduces to the form
$$3 \cdot 668^{2005} = x_{n}^{2} + y_{n}^{2}$$
But again, this equation is possible only if $x_{n}$ and $y_{n}$ are divisible by $3$, and hence $x_{n}^{2} + y_{n}^{2}$ is divisible by $9$. However, since $3 \cdot 668^{2005}$ is not divisible by $9$, such an equation has no solutions.