As Chris Eagle said, $\ker T=\ker L_1\cap \ker L_2$, where $L_1$ and $L_2$ are two linear continuous functionals on $\mathcal C([0,1])$ defined by $L_1(f)=\int_0^1f(t)dt$ and $L_2(f)=\int_0^1tf(t)dt$. We have
$$\operatorname{rank}(T)=\left\{x\mapsto ax+b,a,b\in\mathbb C\right\}.$$
Indeed, the formula $T(f)(x)=x\int_0^1f(t)dt-\int_0^1tf(t)dt$ show that each element of the range should have this form. Since $T(2)(x)=2x-1$ and $T(6X)(x)=3x-2$, we can get all these functions.
For the spectrum, take $\lambda\neq 0$. An eigenvector should have the form $f(t)=at+b$ with $a,b\in\mathbb C$. We get the system
$$\left\{\begin{array}{cc}
\left(\lambda-\frac 12\right)a-b&=0\\\
\frac a3+\left(\lambda+\frac 12\right)b&=0,
\end{array}
\right.
$$
which have a non-trivial solution if and only if $\lambda^2-\frac 14+\frac 13=0$ i.e. $\lambda^2+\frac 1{12}=0$, so $\lambda=\pm\frac i{2\sqrt 3}$. We can check that for these $\lambda$, we have indeed an eigenvector, namely $f(x)=2\sqrt 3x-\sqrt 3\pm i$ for example. Since $T$ is compact in a infinite dimensional vector space, the spectrum is $\{0\}\cup \{\mbox{eigenvalues}\}$, so
$\sigma(T)=\left\{0,\frac i{2\sqrt 3},-\frac i{2\sqrt 3}\right\}$.