1. Find all functions $f: \mathbf{Q}_{+} \rightarrow \mathbf{Q}_{+}$ such that for all $x, y \in \mathbf{Q}_{+}$, we have
$$
f\left(x^{2} f^{2}(y)\right)=f^{2}(x) f(y) .
$$
For all $x \in \mathbf{Q}_{+}$, we have $f(x)=1$.
For any $a, b \in \mathbf{Q}_{+}$, by substituting $x=f(a), y=b$ and $x=f(b), y=a$ into equation (1), we get
$$
\begin{array}{l}
f^{2}(f(a)) f(b) \\
=f\left(f^{2}(a) f^{2}(b)\right) \\
=f^{2}(f(b)) f(a),
\end{array}
$$

which means for all $a, b \in \mathbf{Q}_{+}$, we have
$$
\frac{f^{2}(f(a))}{f(a)}=\frac{f^{2}(f(b))}{f(b)}.
$$

This indicates that there exists a constant $c \in \mathbf{Q}_{+}$ such that:
For all $a \in \mathbf{Q}_{+}$, we have
$$
f^{2}(f(a))=c f(a),
$$

which means for all $a \in \mathbf{Q}_{+}$, we have
$$
\begin{array}{l}
\left(\frac{f(f(a))}{c}\right)^{2}=\frac{f(a)}{c}. \\
\text { Let } f_{n}(x)=\underbrace{f(f(\cdots f}_{n \uparrow}(x) \cdots)) \text { be the } n \text {th }
\end{array}
$$

iteration of $f$.
From equation (2), for any positive integer $n$, we have
$$
\frac{f(a)}{c}=\left(\frac{f_{2}(a)}{c}\right)^{2}=\left(\frac{f_{3}(a)}{c}\right)^{4}=\cdots=\left(\frac{f_{n+1}(a)}{c}\right)^{2 n}.
$$

This indicates that $\frac{f(a)}{c}$ is equal to a $2^n$-th power of a rational number, where $n$ is any positive integer.

If $\frac{f(a)}{c} \neq 1$, then in the prime factorization of $\frac{f(a)}{c}$, the power of some prime cannot be divided by the $2^n$-th power of that prime (for sufficiently large $n$).
Thus, for all $a \in \mathbf{Q}_{+}$, we have
$$
f(a)=c.
$$

Substituting $f(x)=c$ into equation (1) gives
$$
c=c^{3} \Rightarrow c=1.
$$

Therefore, $f(x)=1$ is the unique solution to the original equation (1).