函数 $f(x)=\frac{a x+b}{1+x^{2}}$ 是定义在 $(-1,1)$ 上的奇函数, 且 $f\left(\frac{1}{2}\right)=\frac{2}{5}$.

(1) 确定函数 $f(x)$ 的解析式;

(2)用定义证明: $f(x)$ 在 $(-1,1)$ 上是增函数;

(3)解不等式: $f(t-1)+f(t)<0$.
(1) 由题意知 $\left\{\begin{array}{l}f(0)=0, \\ f\left(\frac{1}{2}\right)=\frac{2}{5},\end{array} \quad\right.$ 即 $\left\{\begin{array}{l}\frac{b}{1+0^{2}}=0, \\ \frac{a}{2}+b \\ 1+\frac{1}{4}=\frac{2}{5},\end{array} \quad\right.$ 解得 $\left\{\begin{array}{l}a=1, \\ b=0,\end{array}\right.$

$\therefore f(x)=\frac{x}{1+x^{2}}$.

(2)证明: 任取 $-1<x_{1}<x_{2}<1$, 则 $x_{2}-x_{1}>0$,

$$
\begin{aligned}
& f\left(x_{2}\right)-f\left(x_{1}\right)=\frac{x_{2}}{1+x_{2}^{2}}-\frac{x_{1}}{1+x_{1}^{2}}=\frac{\left(x_{2}-x_{1}\right)\left(1-x_{1} x_{2}\right)}{\left(1+x_{1}^{2}\right)\left(1+x_{2}^{2}\right)} . \\
& \because-1<x_{1}<x_{2}<1, \\
& \therefore-1<x_{1} x_{2}<1,1-x_{1} x_{2}>0 . \\
& \text { 于是 } f\left(x_{2}\right)-f\left(x_{1}\right)>0, \\
& \therefore f(x) \text { 为 }(-1,1) \text { 上的增函数. } \\
& (3) f(t-1)<-f(t)=f(-t) . \\
& \because f(x) \text { 在 }(-1,1) \text { 上是增函数, } \\
& \therefore-1<t-1<-t<1, \text { 解得 } 0<t<\frac{1}{2} .
\end{aligned}
$$