Bulb A has a volume of 2.0 L and holds nitrogen at 3.0 bar and 300 K…

Step 1. Nitrogen. $n_{\text{N}_2} = \dfrac{3.0 \times 2.0}{0.08314 \times 300} = 0.2406\ \text{mol}$.

Step 2. Helium. The helium starts at $450\ \text{K}$. $n_{\text{He}} = \dfrac{1.0 \times 3.0}{0.08314 \times 450} = 0.0802\ \text{mol}$.

Step 3. The final state. The total amount is $0.2406 + 0.0802 = 0.3208\ \text{mol}$. $P = \dfrac{nRT}{V} = \dfrac{0.3208 \times 0.08314 \times 300}{5.0} = 1.60\ \text{bar}$.

Step 4. Partial pressures. $P_{\text{N}_2} = \dfrac{0.2406 \times 0.08314 \times 300}{5.0} = 1.20\ \text{bar}$ and $P_{\text{He}} = 1.60 - 1.20 = 0.400\ \text{bar}$.