General remarks.
Let $\delta W$ denote the differential work done by a system, so $\delta W$ is postive when the system does work on something else and negative when work is done by something else on the system. For a given process taking place over a path $\gamma$ in thermodynamic state space, the systematic way of determining whether work was done by or on the system is to determine the sign of $W$, the total work done by the system, which is given by
$$
W = \int_\gamma\delta W
$$
This can be computed in various ways depending on the system at hand, and the process it undergoes. The trick is to attempt to find an expression for $\delta W$ that allows for the efficient calculation of the integral for $W$.
Example - adiabatic compression.
Suppose,for example, that we want to determine the work done by the gas during process $1$ of your diagram. Recall that the first law of thermodynamics in differential form can be written as follows:
$$
dE = \delta Q - \delta W
$$
The sign convention here is that $\delta Q$ denotes the heat transferred to the system, and $\delta W$, again, denotes the work done by the system. Since process $1$ is adiabatic, we have $\delta Q = 0$ by definition. It follows that
$$
W = -\int_\gamma dE = -\Delta_\gamma E
$$
where $\Delta_\gamma E$ denotes the total change in energy along the path $\gamma$. Let process $1$ start at point $a$ and end at point $b$, then we can write this result as
$$
W = -(E_b - E_a) = E_a-E_b
$$
So to determine the sign of the work done, we simply need to know whether or not the internal energy of the gas increased (in which case $W<0$ so that work was done on the gas) or decreased (in which case $W>0$ so that work was done by the gas). How to we figure this out for this adiabatic process? Well take, for example, a monatomic ideal gas, and recall that for such a process, we have
$$
T_aV_a^{\gamma-1} = T_bV_b^{\gamma-1}, \qquad \gamma = \frac{5}{3}
$$
Then we see that since $V_b<V_a$, we have $T_b>T_a$; the temperature of the gas increased. But for a monatomic ideal gas, the internal energy can be written purely as a function of temperature and number of particles;
$$
E = \frac{3}{2}NkT
$$
so assuming the number of particles is fixed, the internal energy also increase, and therefore, $W<0$, so work was done on the gas.
