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Answer 1

$f(x)=\frac{a}{\sqrt{a^2-x^2}} \; ,\; \; |x|<a\; \; \; ;\; \; \; f(x)=0\; ,\; \; |x|\geq a\\\\\int\limits^{+\infty }_{-\infty }\, f(x)\, dx=1\; \; \; \Rightarrow \\\\\int\limits^{+\infty }_{-\infty }\, \frac{a\, dx}{\sqrt{a^2-x^2}}=a\cdot \Big (\int\limits^{-a}_{-\infty } \, 0\cdot dx+\int\limits^a_{-a}\, \frac{dx}{\sqrt{a^2-x^2}}+\int\limits_{a}^{+\infty }\, 0\cdot dx\Big )=\\\\=a\cdot arcsin\frac{x}{a}\Big |_{-a}^{a}=a\cdot (arcsin1+arcsin1)=a\cdot 2\cdot \frac{\pi }{2}=\pi \cdot a=1\; ,\\\\a=\frac{1}{\pi}\\\\f(x)=\frac{\frac{1}{\pi }}{\sqrt{\frac{1}{\pi ^2}-x^2}}=\frac{1}{\sqrt{1-(\pi x)^2}}$

$P(\frac{a}{2}<X<a)=P(\frac{1}{2\pi }<X<\frac{1}{\pi })=\int\limits^{\frac{1}{\pi }}_{\frac{1}{2\pi }}\, \frac{dx}{\sqrt{1-(\pi x)^2}}=\\\\=\frac{1}{\pi }\cdot arcsin(\pi x)\Big |_{\frac{1}{2\pi }}^{\frac{1}{\pi }}=\frac{1}{\pi }\cdot (arcsin1-arcsin\frac{1}{2})=\frac{1}{\pi }\cdot (\frac{\pi }{2}-\frac{\pi}{6})=\frac{1}{2}$