Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$
This answer is incomplete as it does not proof all the steps completely:
Let us start with the partial fraction expansion $$\tag{1}\frac{p}{1+x^p}=- \sum_{j=1}^p\frac{\omega^{2j-1}}{x-\omega^{2j-1}}$$ with $\omega= e^{i\pi/p}$.
Now, we can integrate and obtain $$\int_0^1\!dx \frac{p}{1+x^p} = -\sum_{j=1}^p \omega^{2j-1} \log\left(1-\omega^{1-2j}\right). $$
We can simplify this formula a bit taking the real part. This yields $$\int_0^1\!dx \frac{p}{1+x^p} = \pi\sum_{j=1}^{p} \frac{1-2j+p}{2p} \sin\left(\frac{\pi(1-2j)}{p}\right)+\sum_{j=1}^{p}\cos\left(\frac{\pi(1-2j)}{p}\right) \log\left[2 \sin\left(\frac{\pi(1-2j)}{2p}\right)\right].$$
The first term can be summed explicitly $$\int_0^1\!dx \frac{p}{1+x^p} =\frac{\pi}{2\sin(\pi/p)}+\sum_{j=1}^{p}\cos\left(\frac{\pi(1-2j)}{p}\right) \log\left[2 \sin\left(\frac{\pi(1-2j)}{2p}\right)\right].$$ The second term can be related to the digamma function.
Formula (1) can be proven by multiplying the expression by $1+x^p$. For the right hand side, we obtain $$- \sum_{j=1}^p\omega^{2j-1} \prod_{k\neq j} (x-\omega^{2k-1}) = -x^{p-1}\sum_{j=1}^p\omega^{2j-1}+ x^{p-2} \sum_{j=1}^p\omega^{2j-1} \sum_{k,k'\neq j} \omega^{2k-1}\omega^{2k'-1} - \cdots\\+(-1)^p \sum_{j=1}^p\omega^{2j-1} \prod_{k\neq j} \omega^{2k-1}$$ as $1+x^p=\prod_{k=1}^p (x- \omega^{2j-1})$. The result (1) follows from some not so straightforward combinatorics.
I'm only going to address Question 1.
The expression proposed in Question 1 is true. However, it is a little bit too complicated than necessary. A simpler version of the expression is
$$2p\int_0^1 \frac{dx}{1+x^p} = \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right)$$
From reflection formula, take logarithm and differentiate, we get $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin\pi z} \implies \psi(z) - \psi(1-z) = \pi\cot\pi z$$ This leads to $$\frac{\pi}{\sin z} = \pi\cot\frac{\pi z}{2} - \pi\cot\pi z = \psi\left(\frac{z}{2}\right) - \psi\left(1-\frac{z}{2}\right) - \psi(z) + \psi(1-z) $$ From duplication formula, take logarithm and differentiate, we get $$\Gamma(z)\Gamma\left(z+\frac12\right) = 2^{1-2z}\sqrt{\pi}\Gamma(2z) \implies \psi(z) + \psi\left(z + \frac12\right) = -2\log 2 + 2\psi(2z) $$ Apply these to RHS of Question 1, we can expose RHS to following mess $$ \left[ \color{red}{\psi\left(\frac{1}{2p}\right)} - \psi\left(1 - \frac{1}{2p}\right) - \color{green}{\psi\left(\frac{1}{p}\right)} + \color{blue}{\psi\left(1 - \frac{1}{p}\right)} \right] + \left[ \color{green}{2\psi\left(\frac1p\right)} - \color{red}{\psi\left(\frac{1}{2p}\right)} - \color{magenta}{\psi\left(\frac{p+1}{2p}\right)}\right]\\ - \color{green}{\psi\left(\frac{1}{p}\right)} + \psi\left(\frac{p-1}{2p}\right) + \color{magenta}{\psi\left(\frac{p+1}{2p}\right)} - \color{blue}{\psi\left(\frac{p-1}{p}\right)} $$ After massive cancellation, we can simplify RHS to $$ \psi\left(\frac{p-1}{2p}\right) - \psi\left(1 - \frac{1}{2p}\right) = \psi\left( 1 - \left(\frac{1}{2p} + \frac12\right)\right) - \psi\left(1 - \frac{1}{2p}\right) = \psi\left(\frac{1}{2p} + \frac12\right) - \psi\left(\frac{1}{2p}\right) $$ Recall following expansion of digamma function
$$\psi(z) = \frac{1}{z} + \sum_{n=1}^\infty \left(\frac{1}{z+n} - \frac{1}{n}\right)$$ We find $$\begin{align} \text{RHS} &= \frac{1}{\frac{1}{2p}} - \frac{1}{\frac{1}{2p} + \frac12} + \sum_{n=1}^\infty\left(\frac{1}{\frac{1}{2p}+n} - \frac{1}{\frac{1}{2p} + n + \frac12}\right)\\ &= 2\sum_{n=0}^\infty\frac{(-1)^n}{\frac{1}{p}+n} = 2\sum_{n=0}^\infty\int_0^1 (-1)^n t^{\frac{1}{p}+n-1} dt = 2\int_0^1 \sum_{n=0}^\infty (-1)^n t^{\frac{1}{p}+n-1} dt\\ &= 2 \int_0^1 \frac{t^{\frac{1}{p}-1}}{1+t} dt = 2p\int_0^1 \frac{dx}{1+x^p} = \text{LHS} \end{align} $$