Let $K(t,s)= st$ for $s, t \in [0,1]$. Define the operator $T$ on $C[0,1]$ by
$$ (Tf)(t) = \int_0^1 K(t,s)f(s)ds \,\,\,\,\,(\forall \,f \in C[0,1]). $$
a) For any $\lambda$, describe the Null space(= Kernel) of $T - \lambda I$.
b) Find all values of $\lambda$ such that the equation $$ (T-\lambda I)f=g $$ has a unique solution for every $g \in C[0,1]$.
Also, can you explain what is the interpretation of $g$'s that satisfy the above equality?
If $(T-\lambda I)f=0$, you have $$\tag1 \lambda f(t)=\int_0^1 st f(s)\,ds=t\int_0^1 sf(s)\,ds. $$ So, if $\lambda=0$ then $\ker T$ consists of all those $f$ with $\int_0^1 sf(s)\,ds=0$. When $\lambda\ne0$: if $(1)$ holds for $\lambda\ne0$, we have $$\tag2 f(t)=\phi_\lambda(f)\,t, $$ where $$\tag3\phi_\lambda(f)=\frac1\lambda\,\int_0^1 sf(s)\,ds.$$ Then $$\tag4 \phi_\lambda(f)=\frac1\lambda\,\int_0^1 sf(s)\,ds=\frac1\lambda\int_0^1 s\,\phi_\lambda(f)\,s\,ds=\frac1{3\lambda}\,\phi_\lambda(f). $$ For $f$ to be an eigenvector we need $f\ne0$, which forces $\phi_\lambda(f)\ne0$. Then $3\lambda=1$, and so $\lambda=\tfrac13$ is the only possible nonzero eigenvalue. In summary, $\ker(T-\lambda I)=\{0\}$ when $\lambda\not\in\{0,\tfrac13\}$, and $$ \ker(T-\tfrac13\,I)=\{\alpha g:\ \alpha\in\mathbb R\} $$ where $g(t)=t$.
For the second part, if the equation $(T-\lambda I)f=g$ has a unique solution for all $g\in C[0,1]$, this means that $T-\lambda I$ is invertible. Because $T$ is compact (or, instead of using a more general result, we could just notice that $T$ is rank-one), its spectrum consists only of eigenvalues, and so the equation $(T-\lambda I)f=g$ has a unique solution for all $g$ precisely when $\lambda\not\in\{0,\tfrac13\}$.