Sort those 3 logarithmic values without using calculator

As an alternative and to check recall that

$$\log_b a=\frac{\log_c a}{\log_c b}$$

that is

$$\log_{\frac13}27=\frac{\log_2 27}{\log_2\frac13}=-\frac{\log_2 3^3}{\log_2 3}=-3$$

$$\log_{\frac15}4=\frac{\log_2 4}{\log_2 \frac15}=-\frac{\log_2 2^2}{\log_2 5}=-\frac{2}{\log_2 5}$$

$$\log_{\frac12}5=\frac{\log_2 5}{\log_2 \frac12}=-\frac{\log 5}{\log_2 2}=-\log_2 5$$

therefore as you stated

$$\log_{\frac13}27<\log_{\frac12}5<\log_{\frac15}4$$

For clarity, we take the inverses of the bases, which will reverse the order ($\log_{1/x}y=-\log_x y$).

We have

$$\log_3{27}=3,$$

$$\log_54<1,$$ because $4<5^1$ and $$1<\log_25<3$$

because $2^1<5<2^3$.

The rest is yours.

---

Short answer:

$$\log_54<1<\log_25<3=\log_327$$

and

$$\log_{1/5}4>\log_{1/2}5>\log_{1/3}27.$$