Promotion 1 Let $s, t, u, v \in \mathbf{Q}, s \neq u, t \neq v, \lambda \in \mathbf{R}$, function $f(x)$ satisfies
$$
f(s)=t, f(u)=v,
$$

and for any $a, b \in \mathbf{R}$, there is
$$
f(\lambda a+(1-\lambda) b)=\lambda f(a)+(1-\lambda) f(b) .
$$

Find the value of $f(m) (m \in \mathbf{Q})$.
In fact, in the given conditions, let
$$
\begin{aligned}
a= & \frac{x}{\lambda}, b=0 ; a=0, b=\frac{y}{1-\lambda} . \\
& \text { Then } f(x)=\lambda f\left(\frac{x}{\lambda}\right)+(1-\lambda) f(0) ; \\
& f(y)=\lambda f(0)+(1-\lambda) f\left(\frac{y}{1-\lambda}\right) . \\
& \text { Hence } \lambda f\left(\frac{x}{\lambda}\right)=f(x)-(1-\lambda) f(0), \\
& (1-\lambda) f\left(\frac{y}{1-\lambda}\right)=f(y)-\lambda f(0) .
\end{aligned}
$$

Again, in the given conditions, let $a=\frac{x}{\lambda}, b=\frac{y}{1-\lambda}$.
Then $f(x+y)=\lambda f\left(\frac{x}{\lambda}\right)+(1-\lambda) f\left(\frac{x}{1-\lambda}\right)$
$$
=f(x)+f(y)-f(0) \text {. }
$$

Let $g(x)=f(x)-f(0)$. Then
$$
\begin{array}{l}
g(x+y)=g(x)+g(y), \\
g(1)=f(1)-f(0)=s-f(0) .
\end{array}
$$

By the lemma,
$$
g(x)=x g(1)(x \in \mathbf{Q}) \text {. }
$$

Thus, $g(s)=\operatorname{sg}(1), g(u)=u g(1)$.
Also, $g(s)=f(s)-f(0)=t-f(0)$, $g(u)=f(u)-f(0)=v-f(0)$,
then $\left\{\begin{array}{l}\operatorname{sg}(1)=t-f(0), \\ u g(1)=v-f(0) .\end{array}\right.$
Solving, we get $g(1)=\frac{t-v}{s-u}, f(0)=\frac{t u-s v}{u-s}$.
Thus, $g(x)=x g(1)=\frac{t-v}{s-u} \cdot x$.
Hence, $f(x)=g(x)+f(0)$
$$
=\frac{t-v}{s-u} \cdot x+\frac{t u-s v}{u-s}(x \in \mathbf{Q}) \text {. }
$$

Therefore, $f(m)=\frac{t-v}{s-u} \cdot m+\frac{t u-s v}{u-s}(m \in \mathbf{Q})$.
Extension 2 Suppose $n \geqslant 3(n \in \mathbf{N}), s, t, u, v \in \mathbf{Q}, s \neq$ $u, t \neq v, \lambda_{1}, \lambda_{2}, \cdots, \lambda_{n} \in \mathbf{R}$, and $\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n}=1$, the function $f(x)$ satisfies
$$
f(s)=t, f(u)=v,
$$

and for any $a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}$, we have
$$
\begin{array}{l}
f\left(\lambda_{1} a_{1}+\lambda_{2} a_{2}+\cdots+\lambda_{n} a_{n}\right) \\
=\lambda_{1} f\left(a_{1}\right)+\lambda_{2} f\left(a_{2}\right)+\cdots+\lambda_{n} f\left(a_{n}\right) .
\end{array}
$$

Find the value of $f(m)(m \in \mathbf{Q})$.
Extension 2 seems complex, but in fact, by letting $\lambda_{3}=\lambda_{4}=\cdots=\lambda_{n}=0$, the problem reduces to Extension 1.
Thus, $f(m)=\frac{t-v}{s-u} \cdot m+\frac{t u-s v}{u-s}(m \in \mathbf{Q})$.