I have a doubt for Question 3 ( in image ) ,My attempt is not matching the RHS of the proof as you can see below
So Where did I go wrong? ( I seems that if I put n = n - 1 that might solve the RHS a bit but LHS is any positive integer so I can use a special case )
Your mistake is in the 4th line. When you write $$=(-1)^n\frac{(2n+1)}{2}\frac{(2n+3)}{2}...\frac{5}{2}\frac{3}{2}\frac{1}{2}\Gamma(\frac{1}{2})$$ you imply the sequence of numbers $-n+\frac{1}{2},-n-\frac{1}{2},-n-\frac{3}{2}$ is increasing to $\frac{1}{2}$. In reality the sequence you have is decreasing by $1$ each time, and the first term is negative, so this sequence is never getting to $\frac{1}{2}$.
The first three lines are correct, and we can use them: you wrote $$\Gamma(-n+\frac{1}{2})=(-1)\frac{2n+1}{2}\Gamma(-n-\frac{1}{2})$$
Using this, we get $$(-1)\frac{2}{2n+1}\Gamma(-n+\frac{1}{2})=\Gamma(-n-\frac{1}{2})$$
Or, replacing $n$ with $n-1$
$$\Gamma(-n+\frac{1}{2})=(-1)\frac{2}{2n-1}\Gamma(-(n-1)+\frac{1}{2})$$ so $$\Gamma(-n+\frac{1}{2})=(-1)\frac{2}{2n-1}\Gamma(-(n-1)+\frac{1}{2})$$
$$=(-1)^2\frac{2}{2n-1}\frac{2}{2n-3}\Gamma(-(n-2)+\frac{1}{2})$$
$$=(-1)^3\frac{2}{2n-1}\frac{2}{2n-3}\frac{2}{2n-5}\Gamma(-(n-3)+\frac{1}{2})$$
etc.