Laplace Transform: zeros and corresponding impulse response $h(t)$
For you first question you can use the following
$$
\int_{-\infty}^{\infty} \delta (t-a)\,f(t)\,dt = f(a),
$$
with $f(t)$ any function. In your case those integrals would thus yield the values one and zero respectively.
For your second question I will only considering linear time invariant systems. In that case the impulse response of such a system can only contains a Dirac delta function if the transfer function of that system has a numerator of the same order as the denominator. Namely, any transfer function of the form
$$
G(s) = \frac{b_n\,s^n + b_{n-1}\,s^{n-1} + \cdots + b_1\,s + b_0}{s^n + a_{n-1}\,s^{n-1} + \cdots + a_1\,s + a_0},
$$
with $b_n \neq 0$ can also be written as
$$
G(s) = b_n + \frac{b'_{n-1}\,s^{n-1} + \cdots + b'_1\,s + b'_0}{s^n + a_{n-1}\,s^{n-1} + \cdots + a_1\,s + a_0},
$$
with $b'_k = b_k - b_n\,a_k$. The inverse Laplace transform of the constant $b_n$ would contribute a Dirac delta term. For the remaining part of the transfer function one could use partial fraction expansion to show that it can't contribute a Dirac delta term.
If a physical system would have a numerator of the same order as the denominator then it would require that the output of the system is directly affected by the input. An example of such physical system might be some electrical motor where you input a voltage and measure the angular position with some voltage leakage from the input signal to the output. However, most physical systems have numerator of a lower order as the denominator. It is more likely that you might encounter equal order numerators and denominators in digital filters (though, those would be z-domain and not s-domain, but roughly the same argument holds) such as notch filters. Those filters are however often used in series with physical systems, so their combined transfer function would also have a lower order numerator.