44
Since $2025 = 3^4 \times 5^2$, each of $a$, $b$, $c$, and $d$ has no prime factors other than $3$ and $5$. Therefore, we can write
$$a = 3^{x_1} 5^{y_1}, \quad b = 3^{x_2} 5^{y_2}, \quad c = 3^{x_3} 5^{y_3}, \quad d = 3^{x_4} 5^{y_4}$$
for some non-negative integers $x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4$. Since $abcd = 2025$, we obtain
$$x_1 + x_2 + x_3 + x_4 = 4, \quad y_1 + y_2 + y_3 + y_4 = 2.$$

Also, since $ab$, $bc$, $cd$, and $da$ are all perfect squares, the following sums must be even:
$$x_1 + x_2, \quad x_2 + x_3, \quad x_3 + x_4, \quad x_4 + x_1, \\ y_1 + y_2, \quad y_2 + y_3, \quad y_3 + y_4, \quad y_4 + y_1.$$
Thus, all of $x_1, x_2, x_3, x_4$ must have the same parity, and similarly, all of $y_1, y_2, y_3, y_4$ must have the same parity. Conversely, the tuple $(a, b, c, d)$ corresponding to any tuple $(x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4)$ of non-negative integers satisfying (*) and the parity conditions satisfies the given conditions.
Therefore, the possible $(x_1, x_2, x_3, x_4)$ are the permutations of $(4, 0, 0, 0)$, $(2, 2, 0, 0)$, $(1, 1, 1, 1)$, and the possible $(y_1, y_2, y_3, y_4)$ are the permutations of $(2, 0, 0, 0)$. Hence, the number of tuples $(a, b, c, d)$ that satisfy the given conditions is $((\binom{4}{1} + \binom{4}{2}) + 1) \times \binom{4}{1} = 44$.