As a consequence of Euler's refectinreflection and Gauss's multiplication formulas, all values $\Gamma(a)$ with $24a\in\mathbb{Z}$ or $60a\in\mathbb{Z}$ can be expressed algebraically in terms of these values: $$\Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac13\right), \quad \Gamma\!\left(\frac14\right), \quad \Gamma\!\left(\frac15\right), \quad \Gamma\!\left(\frac25\right), \quad \Gamma\!\left(\frac18\right),$$$$ \Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac13\right), \quad \Gamma\!\left(\frac14\right), \quad \Gamma\!\left(\frac15\right), \quad \Gamma\!\left(\frac25\right), \quad \Gamma\!\left(\frac18\right), $$ $$\Gamma\!\left(\frac1{15}\right),\quad \Gamma\!\left(\frac1{20}\right), \quad \Gamma\!\left(\frac1{24}\right), \quad \Gamma\!\left(\frac1{60}\right), \quad \Gamma\!\left(\frac7{60}\right).$$$$ \Gamma\!\left(\frac1{15}\right),\quad \Gamma\!\left(\frac1{20}\right), \quad \Gamma\!\left(\frac1{24}\right), \quad \Gamma\!\left(\frac1{60}\right), \quad \Gamma\!\left(\frac7{60}\right). $$
This is worked out in this paper,in
- Raimundas VIDUNAS, Expressions for the values of the Gamma function, Kyushu Journal of Mathematics, 2005, Volume 59, Issue 2 (2005) Pages 267-283 https://doi.org/10.2206/kyushujm.59.267, arXiv:math/0403510,
along with basic expressions for $a\in(0,1)$. For example, $$\Gamma\!\left(\frac7{10}\right)=\sqrt{\pi}\,2^{3/5}\,\Gamma\left(\frac15\right)^{\!-1}\,\Gamma\!\left(\frac2{5}\right),$$$$ \Gamma\!\left(\frac7{10}\right)=\sqrt{\pi}\,2^{3/5}\,\Gamma\left(\frac15\right)^{\!-1}\,\Gamma\!\left(\frac2{5}\right), $$ $$\Gamma\!\left(\frac1{12}\right)=\frac{3^{3/8}\,\sqrt{\sqrt{3}+1}}{\sqrt{\pi}\,2^{1/4}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac14\right),$$$$ \Gamma\!\left(\frac1{12}\right)=\frac{3^{3/8}\,\sqrt{\sqrt{3}+1}}{\sqrt{\pi}\,2^{1/4}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac14\right), $$ $$\Gamma\!\left(\frac{11}{15}\right)=2\pi\cdot 3^{3/10}\,\Gamma\!\left(\frac15\right)\,\Gamma\left(\frac25\right)^{\!-1}\,\Gamma\left(\frac1{15}\right)^{\!-1},$$$$ \Gamma\!\left(\frac{11}{15}\right)=2\pi\cdot 3^{3/10}\,\Gamma\!\left(\frac15\right)\,\Gamma\left(\frac25\right)^{\!-1}\,\Gamma\left(\frac1{15}\right)^{\!-1}, $$ $$\Gamma\!\left(\frac{11}{20}\right)=2^{1/5}\,\sqrt{5+\sqrt5}\;\Gamma\!\left(\frac15\right)\,\Gamma\!\left(\frac25\right)\,\Gamma\left(\frac1{20}\right)^{\!-1},$$$$ \Gamma\!\left(\frac{11}{20}\right)=2^{1/5}\,\sqrt{5+\sqrt5}\;\Gamma\!\left(\frac15\right)\,\Gamma\!\left(\frac25\right)\,\Gamma\left(\frac1{20}\right)^{\!-1}, $$ $$\Gamma\!\left(\frac{49}{60}\right)=\frac{\sqrt{\pi}\,\sqrt3\,\sqrt{\sqrt3+1}\,\sqrt{5+\sqrt5}\,\sqrt{\sqrt5+\sqrt3}}{5^{1/24}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac1{60}\right)^{\!-1}.$$ Further$$ \Gamma\!\left(\frac{49}{60}\right)=\frac{\sqrt{\pi}\,\sqrt3\,\sqrt{\sqrt3+1}\,\sqrt{5+\sqrt5}\,\sqrt{\sqrt5+\sqrt3}}{5^{1/24}}\,\Gamma\!\left(\frac13\right)\,\Gamma\!\left(\frac1{60}\right)^{\!-1}. $$ Further algebraic independence of the listed 11 "basic" values (and more general answers) are yet to be settled.