Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Then write an explicit expression for (a) $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$, and (b) an expression for $J_1+2J_3$ in terms of $J_0$.
Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$.
Remember that
$J_k(z)=c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz$ where $c_k=\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}$
$J_{k-1}(z)=c_{k-1} z^{k-1} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{3}{2}}dz$ where $c_{k-1}=\frac{1}{2^{k-1} \Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}$
$$\frac{d}{dz}(z^kJ_k(z)) =\frac{d}{dz}(z^k c_k z^k \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k \frac{d}{dz}(z^{2k} \int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}\,dz) = c_k [2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)+ z^{2k} (e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1] $$
But $e^{izs}(1-z^2)^{k-\frac{1}{2}}dz)]_{-1}^1=0$. So 
$$\frac{d}{dz}(z^kJ_k(z)) =c_k 2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz) =\frac{1}{2^k \Gamma(k+\frac{1}{2})\Gamma(\frac{1}{2})}2kz^{2k-1}\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =\frac{1}{2^k-1 (k-\frac{1}{2})\Gamma(k-\frac{1}{2})\Gamma(\frac{1}{2})}kz^{k-1}z^k\int_{-1}^1 e^{izs}(1-z^2)^{k-\frac{1}{2}}dz =z^k c_{k-1}kz^{k-1}\int_{-1}^1 \frac{e^{izs}(1-z^2)^{k-\frac{1}{2}}}{k-\frac{1}{2}}dz$$
How do I contunie from here?
Write an explicit expression for $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$
We know that $J_\frac{1}{2}(x)=\sqrt{\frac{2}{\pi}}\frac{\sin(x)}{\sqrt{x}}$ when $c_\frac{1}{2}=\frac{1}{\sqrt{2}\Gamma(1)\Gamma(\frac{1}{2})}$. By the formula in the previous part, we now that 
$$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x) \iff x^kJ_k(x)=\int_\textbf{R} x^kJ_{k-1}(x)dx \iff J_k(x)=\int_\textbf{R} J_{k-1}(x)dz$$
$$J_\frac{3}{2}(z) =\int_\textbf{R}J_\frac{1}{2}(x) \,dx= \sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx =\sqrt{\frac{2}{\pi}}\int_\textbf{R}\frac{\sin(x)}{\sqrt{x}} dx=(1+i)\frac{\pi}{2}$$
$$J_\frac{5}{2}(x) =\int_\textbf{R}J_\frac{3}{2}(x) \,dx =\int_\textbf{R}(1+i){\frac{\pi}{2}} \,dx $$
Does $J_\frac{5}{2}(x)=\infty$?
Write an explicit expression for $J_1+2J_3$ in terms of $J_0$
By the formula in the previous part,
$$J_1(x) =\int_\textbf{R} J_0(x)\,dx$$
$$J_3(x) =\int_\textbf{R}J_2(x)=\int_{\textbf{R}^2} J_1(x)\,dx^2=\int_{\textbf{R}^3} J_0(x)\,dx^3 $$
So $J_1+2J_3=\int_\textbf{R} J_0(x)\,dx+2\int_{\textbf{R}^3} J_0(x)\,dx^3$
Please check

This is only a partial answer and would be far too long for a comment, but I must quote which is limited in comments.
How are x and z related? Are the $J_k$ the Bessel functions of the first kind? 
For them it is correct that $$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x)$$ 
see http://dlmf.nist.gov/10.6#ii or Abramowitz and Stegun 9.1.30.
So I guess you just mix-up $z$ and $x.\;$ Even with these assumptions your last statements

By the formula in the first part,
$$xJ_1(x)=x^2J_0(x) \iff J_1(x)=xJ_0(x)$$
  $$x^3J_3(x)=x^6J_2(x)=x^9J_1(x)=x^{12}J_0(x) \iff J_3(x)=x^9J_0(x)$$

are blatantly wrong because you obviously omit the derivatives or product rule! For example with $k=0:$ 
$$xJ_0(x)= \frac{d}{dx}(x^1J_1(x))=J_1(x) + xJ_1'(x)$$ 
And if you know a bit about Bessel functions you would immediately see your relations cannot be correct (all higher order  functions would vanish at the zeroes of $J_0$!)
