I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.
1. Why self-inductance is not considered when solving Faraday's law problems
Self inductance should be considered, but is left out for simplicity. So for example, if you have a planar circuit with inductance $L$, resistance $R$, area $A$, and there is a magnetic field of strength $B$ normal to the plane of the circuit, then the EMF is given by $\mathcal{E}=-L \dot{I} - A \dot{B}$.
This means, for example, that if $\dot{B}$ is constant, then, setting $IR=\mathcal{E}$, we find $\dot{I} = -\frac{R}{L} I - \frac{A}{L} \dot{B}$. If the current is $0$ at $t=0$, then for $t>0$ the current is given by $I(t)=-\frac{A}{R} \dot{B} \left(1-\exp(\frac{-t}{L/R}) \right)$. At very late times $t \gg \frac{L}{R}$, the current is $-\frac{A \dot{B}}{R}$, as you would find by ignoring the inductance. However, at early times, the inductance prevents a sudden jump of the current to this value, so there is a factor of $1-\exp(\frac{-t}{L/R})$, which causes a smooth increase in the current.
2. Why an EMF can ever produce a current in a circuit with non-zero self-inductance.
You are worried that EMF caused by the circuit's inductance will prevent any current from flowing. Consider the planar circuit as in part one, and suppose there is a external emf $V$ applied to the circuit (and no longer any external magnetic field). The easiest way to see that current will flow is by making an analogy with classical mechanics: the current $I$ is analogous to a velocty $v$; the resistance is analogous to a drag term, since it represents dissipation; the inductance is like mass, since the inductance opposes a change in the current the same way a mass opposes a change in velocity; and the EMF $V$ is analogous to a force. Now you have no problem believing that if you push on an object in a viscous fluid it will start moving, so you should have no problem believing that a current will start to flow.
To analyze the math, all we have to do is replace $-A \dot{B}$ by $V$ in our previous equations, we find the current is $I(t) = \frac{V}{R} \left(1-\exp(\frac{-t}{L/R}) \right)$, so as before the current increases smoothly from $0$ to its value $\frac{V}{R}$ at $t=\infty$.
Faraday's induction law, that the EMF is proportional to the negative time derivative of the magnetic flux, actually describes two different effects which can be described by the same law. The first effect is based on the flux change due to the change of area of a conducting loop. In this case, the EMF is a purely magnetic force, the magnetic Lorentz force on the moving charge carriers in the conducting loop. The second effect is the induction of an electric field according to the Maxwell-Faraday equation of Maxwell's equations by a changing magnetic flux in a (not necessarily conducting) loop that is not moving relative to the magnetic field . It is, indeed, surprising that both effects can be described by the same induction law.
Best Answer
Let's call $B$ the value of the magnetic field, and let's assume that
$$ B(t) = B(0) -\alpha t$$
where, here, $B(0) = 10\,T$ and $\alpha = 4\,T\cdot s^{-1}$. Then the flux of $B$ through the coil, whose area is $A = 2.5\,m^2$, is
$$\Phi(t) = A(B(0) - \alpha t)$$
Then, Faraday's law tells us that, with the appropriate orientation, this causes an electromotive force $e$ where
$$ e = -\frac{\textrm{d}\Phi}{\textrm{d}t} = A\alpha$$
If you only know that $B(0) = 10\,T$ and $B(t=5\,s)=-10\,T$, then
$$ \langle e \rangle = \frac{1}{T} \int\limits_0^T e\,\textrm{d} t = \frac{1}{T}\left[ -\Phi\right]_0^T = \frac{A(B(0) - B(T))}{T}$$
In both cases, the result is
$$ \langle e \rangle = 10\,V$$