Say we have a sample with a bunch of the original stuff, and you can see when each of them decays. One of them will take $t_1$, another $t_2$, another... If we calculate the average of those, we get a time called 'average life', usually denoted by the greek letter $\tau$:
$\tau = \frac{\sum_i t_i}{N}$
If the sample is big enough, the number of decays we see in a time interval $dt$ is (*)
$ N\frac{dt}{\tau} $,
or, in terms of change in the number of undecayed nuclei,
$ dN=-N \frac{dt}{\tau} $.
If you solve the differential equation, you get:
$ \frac{dN}{dt} = -\frac{N}{\tau} \Rightarrow N(t)=N_0 e^{-t/\tau} $
Which, comparing to yours,
$ N(t)=N_0 e^{-t/\tau}=N_0 e^{-\lambda t} \Rightarrow \lambda = \frac{1}{\tau} $
(*) I know this is overly simplified, but you would probably have to work from Poisson statistics up to really justify this.
Edit: to better answer question 1, let's have a look at the activity by derivating the number of particles not yet disintegrated, $ N(t) $:
$ A(t) = - \frac{dN}{dt} = - \frac{d}{dt} \left( N_0 e^{-\lambda t} \right) = \lambda N_0 e^{-\lambda t} = \lambda N(t) $
So $ \lambda $, besides being the reciprocal of the average life, also links activity and number of radioactive particles.