Let $$I = \int_{-\infty}^{+\infty}f(u)\delta(t-u)\delta(t-2)du$$Find value of $I$.
First of all I don't know if this expression is meaningful in the sense of distributions. The only way which comes to my mind in order to simplify the integral is $$\int_{-\infty}^{+\infty}f(u)\delta(t-u)\delta(t-2)du = \delta(t-2)\int_{-\infty}^{+\infty}f(u)\delta(t-u)du = \delta(t-2)f(t) = f(2)\delta(t-2)$$ Which is by no means a proof. So the main question here is the meaning of $I$ in the sense of distributions.
It can have a meaning in Schwartz' theory of distribution if you are considering the tensor product of two Dirac deltas.
$\delta_t(u):= \delta(u-t)$. Then, $$\langle \delta_t, \psi\rangle = \int \delta_t (u) \psi(u)du = \psi(t)$$ for any $\psi \in \mathcal{D}(\mathbb{R}).$ Similarly; $\delta_2(t):= \delta(t-2)$, then: $$\langle \delta_2, \psi\rangle = \int \delta_2 (t) \psi(t)dt = \psi(2)$$
One can show, that for any $f\in \mathcal{D}(\mathbb{R} \times \mathbb{R})$ the functional: $$ f \to \langle \delta_2, \langle \delta_t, f(u,.)\rangle \rangle $$ is a well defined distribution denoted by $\delta_2 \otimes \delta_t$. $\langle \delta_2, \langle \delta_t, f(u,.)\rangle \rangle = f(t,2)$. (I), without other "integration" over t variable, has no meaning in Schwartz' theory.