= Solution
The <spectrum of a bounded operator> is $\sigma(T)=\{\lambda:\lambda I-T\text{ has no bounded two-sided inverse}\}$. Its <point spectrum> is the set of eigenvalues, where $\lambda I-T$ is not <injective>.
Here $T$ is linear and takes <continuous> functions to continuously differentiable functions. Also $\|Tx\|_\infty\leq2\|x\|_\infty$; the constant function one attains this bound at $t=1$, so $\|T\|=2$. Its range is contained in $C^1([0,1])$, a proper subset of $C([0,1])$, so zero is in its spectrum. It is not an <eigenvalue>: $Tx=0$ gives $x(0)=0$ at zero and $x=0$ upon differentiation.
For a nonzero <eigenvalue>, $\lambda x=Tx$ implies $x'=x/\lambda$ and $(\lambda-1)x(0)=0$. A nonzero solution is therefore possible exactly at $\lambda=1$, with eigenfunctions $x(t)=ce^t$. Hence $p\sigma(T)=\{1\}$.
For the proposed <resolvent> put
$$
F(t)=e^{t/\lambda}\int_0^t y(s)e^{-s/\lambda}ds,\qquad F'=y+F/\lambda.
$$
The proposed $x$ is $y/\lambda+y(0)e^{t/\lambda}/[\lambda(\lambda-1)]+F/\lambda^2$. Consequently $x(0)=y(0)/(\lambda-1)$ and
$$
\int_0^tx(s)ds=\frac{F(t)}\lambda+\frac{y(0)}{\lambda-1}(e^{t/\lambda}-1).
$$
Adding $x(0)$ and subtracting from $\lambda x(t)$ gives exactly $y(t)$. Thus $\lambda I-T$ is onto for $\lambda\ne0,1$. It is also <injective> by the <eigenvalue> calculation. The displayed formula defines a bounded map in the <supremum norm>, since its exponential kernels are bounded on the <compact> interval. It is therefore the <bounded inverse>, and
$$
\boxed{\sigma(T)=\{0,1\},\qquad p\sigma(T)=\{1\}.}
$$
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