Consider the operator
$$ Tu(x)=\int^1_{-1} (1-|x-y|)u(y)dy $$
We want to find the spectrum of $T$.
The kernel is certainly bounded and so this operator is Hilbert-Schmidt, so $T$ is compact. We also notice that the kernel is symmetric, so $T=T^*$. Together these mean the following:
$0\in \sigma_p(T)\cup \sigma_c(T)$
$\sigma_r(T)=\emptyset$
$\lambda\in \sigma(T) , \lambda\neq 0 \implies \lambda \in \sigma_p(T)$
$Tu=\lambda u$ amounts to
$$\int^1_{-1} (1-|x-y|)u(y)dy=\int^1_{-1} u(y)dy-\int^1_{-1}|x-y|u(y)dy=\lambda u$$
this is equivalent to
$$\int^1_{-1} u(y)dy-x\int^x_{-1}u(y)dy+\int^x_{-1}yu(y)dy-\int^1_{x}yu(y)dy+x\int^1_{x}u(y)dy=\lambda u(x) \tag{1} $$
Consider $\lambda \neq 0$. Differentiating $(1)$ gives
$$-\int^x_{-1}u(y)dy+(-xu(x))+xu(x)+xu(x)+\int^1_{x}u(y)dy-xu(x)=\lambda u'(x) $$
or
$$-\int^x_{-1}u(y)dy+\int^1_{x}u(y)dy=\lambda u'(x) \tag{2} $$
differentiating again,
$$-u(x)-u(x)=\lambda u''(x)$$
or
$$0=\lambda u''(x)+2u(x) \tag{3}$$
Letting $\mu=\frac{2}{\lambda}$, we have two cases:
$\mu=-\beta^2<0 \implies u(x)=c_1\sinh(\beta x)+c_2\cosh(\beta x)$
$\mu=\beta^2>0 \implies u(x)=c_3\sin(\beta x)+c_4\cos(\beta x)$
However, it is unclear as to what boundary conditions to use.
Consider $\lambda=0$. $(3)$ becomes
$$u(x)=0\tag{4} $$
This seems to me to mean that $0\in \sigma_c(T)$.
So:
Question 1: (Answered Below!) What boundary conditions should I use for $\lambda\neq 0$?
Question 2: Is my conclusion correct for part 2?
Update: As you can see in the comments below, we have discovered that the boundary conditions are $u'(-1)+u'(1)=0$ and $u(-1)+u(1)=0$.
So for the case $u(x)=c_1\sinh(\beta x)+c_2\cosh(\beta x)$,
$$u(-1)+u(1)=c_1\sinh(-\beta)+c_2\cosh(-\beta)+c_1\sinh(\beta)+c_2\cosh(\beta)=0$$
or considering the even/oddness of $\sinh$ and $\cosh$,
$$2c_2\cosh(\beta)=0 \implies c_2=0$$
Now we see $u'(x)=c_1\beta\cosh(\beta x)$$, so
$$u'(-1)+u'(1)=c_1\beta\cosh(-\beta )+c_1\beta\cosh(\beta)=c_1\beta\cosh(\beta)=0\implies c_1=0$$
So $u(x)$ is not an eigenfunction.
Now consider $u(x)=c_3\sin(\beta x)+c_4\cos(\beta x)$
$$u(1)+u(-1)=c_3\sin(\beta)+c_4\cos(\beta)+c_3\sin(-\beta)+c_4\cos(-\beta) $$
So
$$2c_4\cos(\beta)=0$$
Now consider $u'(x)=c_3\beta\cos(\beta x)-c_4\beta \sin(\beta x)$
So $$u'(1)+u'(1)=c_3\beta\cos(\beta )-c_4\beta \sin(\beta )+c_3\beta\cos(-\beta)-c_4\beta \sin(-\beta)$$
or
$$2c_3\beta\cos(\beta )=0$$
If $c_4=0$,then $\beta=\frac{2n-1}{2}\pi ,n\in\mathbb N$. If $c_3=0$, same. So then $u(x)$ is a nontrivial eigenvector.
So $\mu=\beta^2$ and $\lambda=\frac{2}{\mu}$, so
$$\lambda=\frac{2}{\beta^2}=2\frac{2^2}{(2n-1)^2\pi^2}=\frac{8}{(2n-1)^2\pi^2} >0,n\in\mathbb N$$
So the spectral radius of $T$ is
$$r(\lambda)=\max_n \frac{8}{(2n-1)^2\pi^2} = \frac{8}{(2(1)-1)^2\pi^2} =\frac{8}{\pi^2} $$