The fractions of the form $\frac{n}{n+1}$, with $n$ a positive integer are:

$$
\underbrace{\frac{1}{2}}_{n=1} ; \quad \underbrace{\frac{2}{3}}_{n=2} ; \quad \underbrace{\frac{3}{4}}_{n=3} ; \quad \underbrace{\frac{4}{5}}_{n=4} ; \quad \underbrace{\frac{5}{6}}_{n=5} \cdots
$$

Observe that this sequence of fractions is increasing, that is: $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6} \cdots$

To find the fractions of the form $\frac{n}{n+1}$ that are less than $\frac{7}{9}$, we can compare $\frac{4}{5}$ and $\frac{7}{9}$:

$$
\frac{4}{5}=\frac{36}{45} \quad \text{and} \quad \frac{7}{9}=\frac{35}{45}
$$

Therefore, $\frac{4}{5}>\frac{7}{9}$
Thus, 4/5 is greater than 7/9, and since the sequence is increasing, from $4 / 5$ onwards all the fractions in this sequence are greater than 7/9. Therefore, the fractions of the form $\frac{n}{n+1}$ that are less than $\frac{7}{9}$ are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}$. Hence, the answer is 3.
2 - Converting to decimal numbers we have: $7 / 9=0.777 \ldots$ and $1 / 2=0.5$; $2 / 3=0.666 \ldots ; 3 / 4=0.75 ; 4 / 5=0.8 ; 5 / 6=0.8333 \ldots$

Therefore, the sequence is increasing and only $1 / 2=0.5 ; 2 / 3=0.666 \ldots ; 3 / 4=0.75$ are less than $7 / 9=0.777 \ldots$

Solution 3 - If $\frac{n}{n+1}<\frac{7}{9}$, we must have $2 n-7<0$, that is $n<\frac{7}{2}=3.5$. Therefore, $n=1,2,3$ and thus, the fractions are $\frac{1}{2}, \frac{2}{3} \text{ and } \frac{3}{4}$.