Hello, Tram!
Welcome aboard!
Here are a few of them . . .
\(\displaystyle 5)\;\log_5(\sqrt{125})\,+\,32^{-\frac{1}{5}}\)
We have: \(\displaystyle \,\log_5(125)^{\frac{1}{2}}\,+\,(2^5)^{-\frac{1}{5}}\;=\;\frac{1}{2}\cdot\log_5(125)\,+\,2^{(5)(-\frac{1}{5})}\;=\;\frac{1}{2}\cdot\log_5(5^3)\,+\,2^{-1}\)
\(\displaystyle \;\;\;=\;3\cdot\frac{1}{2}\cdot\log_5(5)\,+\,\frac{1}{2^1}\;=\;\frac{3}{2}\cdot1\,+\,\frac{1}{2}\;=\;2\)
\(\displaystyle 8)\;\tan\left(\frac{7\pi}{4}\right)\,+\,\cos\left(\frac{7\pi}{6}\right)\)
You're expected to know the trig values for \(\displaystyle 30^o,\;60^o,\;45^o\) and their variations.
\(\displaystyle \tan\left(\frac{7\pi}{4}\right)\,+\,\cos\left(\frac{7\pi}{6}\right)\;=\;-1\,+\,\left(-\frac{\sqrt{3}}{2}\right)\;= \;-\frac{2\,+\,\sqrt{3}}{2}\)
\(\displaystyle 11)\;\cos\left(\frac{11\pi}{12}\right)\)
Formula: \(\displaystyle \,\cos(A\,+\,B)\;=\;\cos(A)\cdot\cos(B)\,-\,\sin(A)\cdot\sin(B)\)
Since \(\displaystyle \,\frac{11\pi}{12}\:=\:\frac{3\pi}{4}\,+\,\frac{\pi}{6}\)
we have: \(\displaystyle \,\cos\left(\frac{3\pi}{4}\,+\,\frac{\pi}{6}\right)\;=\;\cos\left(\frac{3\pi}{4}\right)\cdot\cos\left(\frac{\pi}{6}\right)\,-\,\sin\left(\frac{3\pi}{4}\right)\cdot\sin\left(\frac{\pi}{6}\right)\)
\(\displaystyle \;\;\;\;= \;\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\,-\,\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) \;= \;-\frac{\sqrt{6}}{4}\,-\,\frac{\sqrt{2}}{4}\;=\;-\frac{\sqrt{6}\,+\,\sqrt{2}}{4}\)