Given the volumes: determine the pH and the final temperature of a mixture knowing only the initial pH and the temperature of the un-mixed components
For the temperature, this method of calculation will give you a rough estimate of the final temperature, but it will not be exactly right. The reason is that mixing two solutions comes at an energy cost/gain due to the interaction of the two solution molecules. For some mixtures this enthalpy of mixing yields a temperature rise (i.e. energy is released from the molecular potential energy), for others it yields a temperature decrease (i.e. energy is taken up to 'make the mixing possible'). For many mixtures this difference is not more than a few degrees, which makes your linear estimate reasonable, but it is good to keep in mind that it will not be exact.

For the pH, the shortcut you are taking is incorrect and will at best give you a very crude estimate. The reason is that the pH scale is logarithmic, it is correlated with the concentration of protons in the solution as $\ce{pH}=-\log_{10}\; [\ce{H+}] $, where $[\ce{H+}]$ is the proton concentration. (I will make the assumption here that you have simple liquids, which are not buffered.) Therefore, you would first need to calculate the proton concentration in both solutions from this equation, yielding $10^{-3}\,\ce{M}$ and $10^{-5}\,\ce{M}$ for sols A and B resp. This means that the amount of $\ce{H+}$ in sol A is $2\cdot10^{-3}\,\ce{mol}$ and that in sol B $10^{-5}\,\ce{mol}$. Mixing the two you get 3 liter of solution with $2.01\cdot10^{-3}\,\ce{mol}$ protons, which is $6.7\cdot10^{-4}\,\ce{M}$ which works out to $\ce{pH=}3.17$. Indeed, very far off from the 3.67 predicted in your linear method.

Short discussion on the pH: because the scale is logarithmic and your 2 $\ce{pH}$ values are actually not that far from each other the linear estimate is still ok-ish. If, for example, you would have used $\ce{pH}$ 0 and 7 you would have found $\ce{pH}=0.17$ vs. $2.33$ from your estimate.