1. Find all functions $f$ satisfying $f(1)=2$ and $f(x y)=f(x) \cdot f(y)-f(x+y)+1, x, y \in \mathbf{Q}$:
$$
\mathbf{Q} \rightarrow \mathbf{Q}
$$
1. In the original functional equation, let $y=1$, we get $f(x) \equiv f(x) \cdot f(1)-f(x+1)+1(x \in \mathbf{Q})$, i.e., $f(x+1) \equiv f(x)+1$. Therefore, for all $x \in \mathbf{Q}, n \in \mathbf{Z}$, we have $f(x+n)=f(x)+n$. Thus, $f(n)=f(1)+n-1=n+1$. Actually, in the original functional equation, taking $x=\frac{1}{n}, y=n, x \in \mathbf{Z}$, we get $f\left(\frac{1}{n} \cdot n\right)=f\left(\frac{1}{n}\right) \cdot f(n)-f\left(\frac{1}{n}+n\right)+1$, thus $2=$ $f\left(\frac{1}{n}\right)(n+1)-f\left(\frac{1}{n}\right)-n+1$, i.e., $f\left(\frac{1}{n}\right)=1+\frac{1}{n}$. Finally, taking $x=p, y=\frac{1}{q}, p \in \mathbf{Z}, q \in \mathbf{N}$, we get $f(p$. $\left.\frac{1}{q}\right)=f(p) \cdot f\left(\frac{1}{q}\right)-f\left(p+\frac{1}{q}\right)+1$, from which we get $f\left(\frac{p}{q}\right)=(p+1)\left(\frac{1}{q}+1\right)-\frac{1}{q}-p=\frac{p}{q}+1$. Therefore, only the function $f(x)=x+1$ satisfies all the conditions given in the problem.