4. Let $f(x)$ be an odd function defined on $\mathbf{R}$, such that for any $x \in \mathbf{R}$, $f(x+2)=f(x)$, and when $x \in(0,1)$, $f(x)=x^{2}$. Then
$$
f\left(-\frac{3}{2}\right)+f(1)=
$$
$\qquad$
4. $\frac{1}{4}$.

From the problem, we know
$$
f\left(-\frac{3}{2}\right)=f\left(-\frac{3}{2}+2\right)=f\left(\frac{1}{2}\right)=\frac{1}{4} \text {. }
$$

Since $f(x)$ is an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have $f(x+2)=f(x)$, therefore,
$$
\begin{array}{l}
f(1)=f(-1)=-f(1), f(1)=0 . \\
\text { Hence } f\left(-\frac{3}{2}\right)+f(1)=\frac{1}{4} .
\end{array}
$$