[Solution] (1) When $x=2$, the denominator of the fraction is 0, which is not appropriate. (1) is not true.
(2) From $(a+1)^{0}=1$, it does not follow that $a \neq 0$, so (2) is not true.
(3) The domain of the function should also exclude $x \neq 0$ (the denominator of the fraction cannot be 0), hence (3) is not true. It is easy to see that (4) is true.
Therefore, the answer is $(A)$.
$7 \cdot 81$ Among the following propositions:
(1) The fraction $\frac{x^{2}-4}{x-2}$, when $x= \pm 2$, its value is zero.
(2) If $(a+1)^{0}=1$, then $a \neq 0$.
(3) In the function $y=\sqrt{x+\frac{1}{2}}+\frac{1}{x}$, the domain of the independent variable is $x \geqslant-\frac{1}{2}$.
(4) In the same coordinate system, the graphs of $y=x+m$ and $y=\frac{m}{x}$ (where $m>0$) are roughly as shown in the right figure.

The number of correct propositions is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(China Guangzhou, Wuhan, Fuzhou, etc. five cities junior high school mathematics competition, 1991)