Hints/guidelines:
$1$. Prove that $\Gamma(z)$ has no zeroes so $\psi(z)$ can have no singular points other than the poles $0,-1,-2,...$. Then it's not too hard to prove using the properties of $\Gamma$ function that $\psi$ has the representation
$$
\psi(z)=-\frac{1}{z+n}+\mathcal{E}(z+n)
$$
Here $\mathcal{E}(z+n)$ is the regular part of $\psi(z)$.
$2$. See the comment.
$3.$ Non-detailed proof:
Using the definition $\Gamma'(z)$ and by replacing the $\text{log}(t)$ term with the integral
$$
\text{log}(t)= \int_0^{\infty}\frac{e^{-x}-e^{-xt}}{x},
dx
$$
we can show that
$$
\Gamma'(z)=\int_0^\infty \frac{dx}{x}\left[ e^{-x}\Gamma(z)-\int_0^\infty e^{-t(x+1)}t^{z-1} dt \right].
$$
Using the substitution $u=t(x+1)$ we get
$$
\psi(z)=\int_0^\infty\left[e^{-x}-\frac{1}{(x+1)^{z}} \right]\frac{dx}{x}.
$$
Using this we can find
$$
\psi(z)=\lim_{\epsilon \to 0} \left[ \int_\epsilon^\infty \frac{e^{-x}}{x}dx-\int_\epsilon^\infty \frac{1}{(x+1)^{z}x} dx \right].
$$
Substituting $1+x = e^u$ for the last integral we get
$$
\psi(z)=\lim_{\epsilon \to 0} \left[ \int_{log(1+\epsilon)}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du -\int_{log(1+\epsilon)}^\epsilon \frac{e^{-u}}{u} du \right].
$$
Here the last integral goes to $0$ as $\epsilon \to 0$ so we get
$$
\psi(z)= \int_{0}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du.
$$
Plugging in $z=1$, subtracting the result and using substitution $x=e^{-u}$ we finally get
$$
\psi(z)=-\gamma+\int_0^{1}\frac{1-x^{z-1}}{1-x}dx.
$$
Now the claim follows by using $(1-x)^{-1}=\sum_{n=0}^{\infty} x^n$ and integrating term by term.
