The short answer is that you're right - the change in entropy due to the decrease in volume cancels out the change in entropy due to the increase in temperature. But you're also right that this involves a simultaneous change in $p$ and $T$, and it would be nice to know by how much each of these changes. Below I've done the calculation. It was a bit more involved than I was expecting, and I hope I didn't make any mistakes!
To do this calculation we first need to note that the ideal gas law,
$$pV=nRT,\tag{i}$$ tells us how pressure, volume and temperature relate to one another, but if you want to do any real calculations, you also need to know how the internal energy $U$ behaves. This is given by
$$
U =n\; c_V T,\tag{ii}
$$
where $c_V$ is the dimensionless heat capacity at constant volume. (Some people will define an ideal gas in such a way that $c_V$ is allowed to be a function of $U$, but here I'll assume it's constant. To a good approximation, $c_V=\frac{3}{2} R$ for a monatomic gas, or $\frac{5}{2} R$ for a diatomic one.) Equation $(\mathrm{ii})$ can't be derived from Equation $(\mathrm{i})$, so both of them are needed in order to define the properties of an ideal gas.
With this in mind, let's start with the fundamental equation of thermodynamics (for systems without chemical reactions):
$$
dU = TdS - pdV.
$$
Because we're considering an adiabatic process we know that $TdS = 0$, so
$$
dU = -pdV = - \frac{nRT}{V} dV,
$$
where the second equality is obtained by substituting the ideal gas law $(\mathrm{i})$. But we also know from the heat capacity equation $(\mathrm{ii})$ that $nT = U/c_V$. This gives us
$$
dU = - \frac{U\;R}{c_VV} dV,
$$
or
$$
c_V\frac{1}{U}dU = - R\frac{1}{V} dV.
$$
Now we can integrate both sides:
$$
c_V \int_{U_1}^{U_2} \frac{1}{U}dU = -R\int_{V_1}^{V_2} \frac{1}{V}dV,
$$
or
$$
c_V\left( \ln U_2 - \ln U_1 \right) =R \left( \ln V_1 - \ln V_2 \right).
$$
A quick note about interpretation is in order here. We're integrating both sides over different ranges ($U_1$ to $U_2$ and $V_1$ to $V_2$) and then setting them equal. This is because we want to know how much the internal energy will change if we reversibly change the volume by a certain amount, so we're looking for $U_2$ and $U_1$ as a function of $V_1$ and $V_2$. Anyway, now we can use some logarithm identities to get
$$
c_V \ln \frac{U_2}{U_1} = R\ln \frac{V_1}{V_2}
$$
or
$$
\ln \left(\frac{U_2}{U_1}\right)^{c_V} = \ln \left(\frac{V_1}{V_2}\right)^R,
$$
so
$$
\left(\frac{U_2}{U_1}\right)^{c_V} = \left(\frac{V_1}{V_2}\right)^{R},
$$
or
$$
V_1^{R} U_1^{c_V} = V_2 ^{R} U_2^{c_V}.
$$
This means that, for a reversible process, the quantity $VU^{c_V}$ must remain constant. Now, finally, we can substitute $U$ from Equation $(\mathrm{ii})$ to get
$$
V^R(c_VnT)^{c_V} = \text{constant for an adiabatic process.}
$$
There's a factor of $(c_Vn)^{c_V}$ that we can ignore by incorporating it into the constant on the right hand side, so
$$
V^RT^{c_V} = \text{constant for an adiabatic process,}
$$
or
$$
V_1^R T_1^{c_V} = V_2^R T_2^{c_V}.
$$
Given any values for $V_1$, $V_2$ and $T_1$, you can use this to work out $T_2$. By substituting into $(\mathrm{i})$ you can also work out the change in pressure.
The total work done by the system in one cycle is the sum of the works done in the isothermal processes:
$$W_{total}=nR\left ( T_h\ln \frac{V_2}{V_1}-T_c\ln\frac{V_3}{V_4} \right )$$
Also for adiabatic processes we have :
$$T_h V_2^{\gamma-1 }=T_c V_3^{\gamma-1}\tag{2}$$
$$T_hV_1^{\gamma-1}=T_cV_4^{\gamma-1}\tag{3}$$
Substituting for $V_2$ and $V_4$ from $(2)$ and $(3)$ in the first equation, we can find the ratio $V_3\over V_1$. (Find the final result yourself!)
Best Answer
By saying that it is insulated they are saying that heat can't enter or leave the system. So you have no change in entropy due to $\Delta S = \Delta Q/T$. So if the entropy is constant throughout the process you can use the reversible adiabatic formula. And yes the temperature and pressure do change as you compress the gas (they increase).
Strictly speaking the 'reversible' part is in question. If there are internal processes such as friction or viscosity, the entropy of the gas may increase. But in that case the work needed would be even higher since all else being equal higher entropy means higher internal energy. But they only ask how much work is "necessary". And even if they didn't say that you could assume that from the context as a homework problem.