4. 

$$
\frac{S_{M N C}}{S_{A B C}}=\frac{(p-c)^{2}}{a b}=\frac{(a+b-c)^{2}}{4 a b}=\frac{4(a+b)^{2}}{81 a b}
$$

Since $a+b \geqslant 2 \sqrt{a b}$ (equality is achieved only when $a=b$), then $\frac{16(a+b)^{2}}{81 a b} \geqslant \frac{4 \cdot 4 a b}{81 a b}=\frac{16}{81}$.

Rewrite the ratio of the areas in the following form:

$$
\frac{S_{M N C}}{S_{A B C}}=\frac{4(a+b)^{2}}{81 a b}=\frac{4}{81}\left(\frac{a}{b}+2+\frac{b}{a}\right)=\frac{4}{81}\left(t+\frac{1}{t}+2\right)
$$

where $t=a / b$. From the condition of the existence of the triangle, the inequalities follow: $a+b>c, a+c>b, b+c>a$, considering that $c=5(a+b) / 9$, the last inequalities are equivalent to the following: $7 / 2>a / b>2 / 7$. Thus $t \in(2 / 7 ; 7 / 2)$. The function $f(t)=t+1 / t$ is monotonically decreasing on $(0 ; 1)$ and increasing on $(1 ;+\infty)$. Since $f(2 / 7)=f(7 / 2)$, then $f(t) \in[f(1) ; f(7 / 2))$ for $t \in(2 / 7 ; 7 / 2)$. Therefore $\frac{S_{M N C}}{S_{A B C}} \in[16 / 81 ; 2 / 7)$.
Answer: a) $16 / 81$. b) $[16 / 81 ; 2 / 7)$. Answer to variant: 2-2: a) $64 / 225$. b) $[64 / 225 ; 4 / 11)$.