Finding values of DTFT without explicitly computing
You just need to know the formulas for the DTFT and its inverse:

$$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{1}$$

and

$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{jn\omega}d\omega\tag{2}$$

From $(1)$ you see that your answer for $(a)$ is correct. Also the answer for $(d)$ follows immediately, if you realize that $e^{-jn\pi}=(-1)^n$.

For $(b)$ it's important to see that the coefficients are asymmetrical. Looking at $(1)$ you see that you can pair the terms for indices $\pm 1$, $\pm 2$, etc., which allows you to rewrite the sum $(1)$ as a sum of weighted sines time $j$ (imaginary unit), because $e^{jx}-e^{-jx}=2j\sin(x)$. So you'll have a purely imaginary expression. Computing the phase should then be easy.

$(c)$ is easily solved using formula $(2)$ (which value of $n$ do you need to plug into $(2)$?). And finally, for $(e)$ you're right that you need Parseval's theorem.