For part (a) finding the cdf $F(w)$ of $W$,my way to do it: $F(v)=F(g(v))= \begin{cases} 1/(5 \sqrt{2}) \int_{-\infty}^{-10} e^{-(-10)^2/50}\, dx & \text{$v<-10$} \\ 1/(5 \sqrt{2}) \int_{-10}^{10} e^{-(v)^2/50}\,dx & \text{$-10≤v≤10$}\\ 1/(5 \sqrt{2}) \int_{10}^{\infty} e^{-(10)^2/50}\, dx & \text{$v≥10$} \end{cases}$
I am not sure about my way to do it is correct. What do you think?
Notice that for $-10\leq t < 10$, $w \leq t$ whenever $v \leq t$. In addition, $w$ is always less than or equal to $10$ and never less than $-10$. So $$P(w \leq t) = 0,\qquad t < -10$$ $$P(w \leq t) = P(v \leq t) = \frac{1}{5\sqrt{2\pi}}\int_{-\infty}^te^{-x^2/50}\;dx,\qquad t< 10$$ $$P(w \leq 10) = 1$$ We may write this in one equation using the step function $u$: \begin{align} F_W(w)&=u(w+10)\left(\frac{1-u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50}\;dx + u(w-10)\right) \\ &=\frac{u(w+10)-u(w+10)u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50}\;dx + u(w+10)u(w-10) \\ &= \frac{u(w+10)-u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50}\;dx + u(w-10) \end{align} We may take the derivative of $F_W$, remembering that $\frac{d}{dx}u(x) = \delta(x)$, where $\delta(x)$ is the Dirac delta: $$f_w(w)=\frac{\delta(w+10) - \delta(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^we^{-x^2/50}\;dx + \frac{u(w+10)-u(w-10)}{5\sqrt{2\pi}}e^{-w^2/50} + \delta(w-10)$$