transformation of single random variables with absolute value ?? 
integral
I got the final answer to be fy(y)= 1   0< y < 1
I am not sure could anyone correct me if its wrong !
 A: Since the the map $1-|x|$ is not invertable on $[-1,1]$, you cannot apply the general formula, you have to do some calculations instead:
a) Integrate over $[-1,y]$ to get the CDF:
$$\int_{-1}^y \frac{1-x}{2}\mathrm{d}x=C+\frac{2y-y^2}{4}$$
As it is a CDF, $C=\frac{3}{4}$, so
$$F_X(x)=\frac{3+2x-x^2}{4}$$
b) $|X|\leq 1$, so:
 $$\mathbb{P}(1-|X|<t)=\mathbb{P}(1-t<|X|)=\mathbb{P}(X<t-1\cup 1-t<X)=F(t-1)+1-F(1-t)=t$$
 $$F_Y(t)=t\Rightarrow f_Y(t)=1$$
 c) $0.6$ as it is a uniform random variable
d) $\mathbb{E}(Y)=0.5$ and 
 $$\int_{-1}^{1}x\frac{1-x}{2}\mathrm{d}x=-\frac{1}{3}$$
A: Yes.  Your answer is correct.  
Ákos Somogyi showed one method to obtain it; by integration, substitution, then derivation.  
Another method is to use the change of variables transformation.
$y=1-\lvert x\rvert$ means $x =\pm(1-y)$
The standard formula is modified because the support of $X$ is subdivided into two intervals which both map to the support of $Y$.
$$\begin{align}
f_Y(y) & = f_X(1-y)\cdot\left\lvert \frac{\mathrm d (1-y)}{\mathrm d y\qquad}\right\rvert+f_X(y-1)\cdot\left\lvert \frac{\mathrm d (y-1)}{\mathrm d y\qquad}\right\rvert
\\[2ex] & = \left[(-0.5(1-y)+0.5)+(-0.5(y-1)+0.5)\right]\cdot\mathbf 1_{y\in[0;1]}
\\[2ex] & = \mathbf 1_{y\in[0;1]}
\end{align}$$
