Two cells of emf 6 V (internal resistance 1 Ω) and 4 V (internal resistance 2 Ω) are connected in parallel, positive terminal to positive terminal…

Let $V$ be the common terminal voltage. Kirchhoff's current law at the positive node gives $\dfrac{6-V}{1}+\dfrac{4-V}{2}=\dfrac V3$, i.e. $8=\dfrac{11}{6}V$, so $V=\dfrac{48}{11}\approx4.36\ \mathrm{V}$. The current in the resistor is $V/3=16/11\ \mathrm{A}\approx1.45\ \mathrm{A}$ (B true) and the potential difference is $4.36\ \mathrm{V}$, not $4\ \mathrm{V}$ (D false). The current in the $6\ \mathrm{V}$ cell is $(6-V)/1\approx1.64\ \mathrm{A}$, while for the $4\ \mathrm{V}$ cell $(4-V)/2\approx-0.18\ \mathrm{A}$ is negative: since $V>4$ the current enters its positive terminal, so it is being charged (A true). The equivalent emf of two parallel cells is $\dfrac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}=\dfrac{8}{1.5}=5.33\ \mathrm{V}$, not $5\ \mathrm{V}$ (C false).