Solution:
Proof of $3 \leq \frac{4 a+b}{a+4 b}+\frac{4 b+c}{b+4 c}+\frac{4 c+a}{c+4 a}$ :

1st variant: Multiplying by the common denominator and simplifying leads to the equivalent inequality $45 a b c \leq 7\left(a^{2} b+b^{2} c+c^{2} a\right)+8\left(a b^{2}+b c^{2}+c a^{2}\right)$. This follows from the inequality between the arithmetic and geometric mean (e.g. $a^{2} b+b^{2} c+c^{2} a \geq 3 \sqrt[3]{a^{2} b \cdot b^{2} c \cdot c^{2} a}=3 a b c$ ).

2nd variant: For $n \geq 1$ and real numbers $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ the Cauchy-Schwarz inequality $\sqrt{a_{1}^{2}+\ldots+a_{n}^{2}} \sqrt{b_{1}^{2}+\ldots+b_{n}^{2}} \geq a_{1} b_{1}+\ldots+a_{n} b_{n}$ holds. For $n=3$, $a_{1}^{2}=\frac{4 a+b}{a+4 b}$, $a_{2}^{2}=\frac{4 b+c}{b+4 c}$, $a_{3}^{2}=\frac{4 c+a}{c+4 a}$, $b_{1}^{2}=(4 a+b)(a+4 b)$, $b_{2}^{2}=(4 b+c)(b+4 c)$, $b_{3}^{2}=(4 c+a)(c+4 a)$ this gives

$$\left(\frac{4 a+b}{a+4 b}+\frac{4 b+c}{b+4 c}+\frac{4 c+a}{c+4 a}\right)((4 a+b)(a+4 b)+(4 b+c)(b+4 c)+(4 c+a)(c+4 a)) \geq (5(a+b+c))^{2}$$

The second factor, because of $2(a+b+c)^{2}-6(a b+b c+c a)=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$, is less than $\frac{25}{3}(a+b+c)^{2}$, from which the claim follows.

Proof of $\frac{4 a+b}{a+4 b}+\frac{4 b+c}{b+4 c}+\frac{4 c+a}{c+4 a}<\frac{33}{4}$ :

1st variant: Multiplying by the common denominator and simplifying gives the equivalent (trivial) inequality $0<127 a b c+16\left(a^{2} b+b^{2} c+c^{2} a\right)$.

2nd variant: The inequality is equivalent to $\frac{b}{a+4 b}+\frac{c}{b+4 c}+\frac{a}{c+4 a}>\frac{1}{4}$; the latter holds because of

$$\frac{b}{a+4 b}+\frac{c}{b+4 c}+\frac{a}{c+4 a}>\frac{b}{4 a+4 b+4 c}+\frac{c}{4 a+4 b+4 c}+\frac{a}{4 a+4 b+4 c}=\frac{1}{4}$$