Let's determine all functions $f$ defined on the set of non-negative real numbers $R_{0}^{+}$, which only take non-negative real values, and satisfy the following three conditions:
a) $f(x \cdot f(y)) \cdot f(y)=f(x+y) \quad$ for all $\quad x, y \in R_{0}^{+} \quad$;
b) $f(2)=0$
c) $f(x) \neq 0$, if $0 \leq x<2$.
5. Task. Determine all functions $f$ defined on the set of non-negative real numbers $R_{0}^{+}$, which only take non-negative real values and satisfy the following three conditions:

(a) $f(x \cdot f(y)) \cdot f(y)=f(x+y) \quad$ for all $\quad x, y \in R_{0}^{+} \quad$;

(b) $f(2)=0$

(c) $f(x) \neq 0, \quad$ if $\quad 0 \leq x < 2$

From (b), we know that $f(2)=0$. From (c), we know that $f(x) \neq 0$ for $0 \leq x < 2$. We will show that for $0 < y < 2$, $f(y) = \frac{2}{2-y}$. Suppose there exists a $y$ such that $0 < y < 2$ and $f(y) \neq \frac{2}{2-y}$. Then, $f(y) < \frac{2}{2-y}$ or $f(y) > \frac{2}{2-y}$. If $f(y) < \frac{2}{2-y}$, then for $x = \frac{2}{f(y)} - y$, we have $x > 0$ and $x + y = \frac{2}{f(y)} < 2$, so $f(x + y) \neq 0$. However, $x \cdot f(y) = 2$, and thus $f(x \cdot f(y)) = f(2) = 0$. This contradicts (a). If $f(y) > \frac{2}{2-y}$, then $f(y) = \frac{2}{2-y} + d$, where $d$ is positive. Then, there exists a positive $x$ such that $\frac{2}{f(y)} = \frac{2}{\frac{2}{2-y} + d} < 2 - y$, or $x \cdot f(y) > 2$, and thus $f(x \cdot f(y)) = 0$. Therefore, (a) does not hold for these values, and this contradiction shows that for all $0 < y < 2$, $f(y) = \frac{2}{2-y}$. Thus, we have determined that the function satisfying the conditions of the problem is:

$$
f(x)= \begin{cases}\frac{2}{2-x}, & \text { if } \quad 0 \leq x < 2 \\ 0, & \text { if } \quad x \geq 2\end{cases}
$$

Now, we only need to check whether this function indeed satisfies the requirements. However, we leave this verification to the reader, as it is easy to see that this function is indeed valid.