Question about differential amplifier
Your circuit is:







simulate this circuit &ndash; Schematic created using CircuitLab

You know that the emitter voltage for both BJTs is the same (they are tied together.) It follows from the schematic that \$\mid\: V_{\text{BE}_2}\mid\:\:=\:\:\mid V_{\text{BE}_1}\mid+500\:\text{mV}\$. You should be able to easily see that fact, directly from the schematic: \$V_{\text{B}_1}=V_{\text{B}_2}+500\:\text{mV}\$ and \$V_{\text{E}_1}=V_{\text{E}_2}\$.

Since you also know that for every \$60\:\text{mV}\$ difference in \$V_\text{BE}\$ there will be about \$10\times\$ the collector current, it follows that \$\frac{I_{\text{C}_2}}{I_{\text{C}_1}}\approx e^{^\frac{500\:\text{mV}}{26\:\text{mV}}}\approx 225\times10^{6}\$. In other words, \$Q_2\$ hogs all available current in \$R_1\$. What remains for \$Q_1\$ is in the small "parts per billion." (In short, nothing at all.)

From here, you have to only decide the approximate collector current for \$Q_2\$. If you temporarily assume that \$V_{\text{BE}_2}=700\:\text{mV}\$ then you find that \$I_{\text{C}_2}=\frac{5\:\text{V}-700\:\text{mV}}{1\:\text{k}\Omega}\approx 4.3\:\text{mA}\$. Since this value is fairly consistent with \$V_{\text{BE}_2}\approx 700\:\text{mV}\$ for most small-signal BJTs, you can reasonably rest on this computed value.

This assumption confirmed, the result is easily worked out.