As stated and illustrated in the comments by Mindlack the first part is clearly wrong. Consider $\rho=\sigma=\tau=0$ the trivial homomorphism. Then $\ker\rho=F_1$ and $\ker\sigma= E_1$ and you are asked to show that $E_1\subset\varphi_1(F_1)$ which is only the case for $\varphi_1$ being surjective. Pick your favorite non-surjective $\varphi_1$ and see that this cannot work. However, the reverse inclusion holds.
To see this, consider $x\in\ker\rho$. Then by commutativity
$$(\varphi_2\circ\rho)(x)=(\sigma\circ\varphi_1)(x)=0$$
and thus $\varphi_1(\ker\rho)\subset\ker\sigma$. The second part is correct. Pick $\tau(x)\in{\rm im}\,\tau$ for some $x\in G_1$. By exactness, $\psi_1$ is surjective and therefore there is some $y\in E_1$ such that $\psi_1(y)=x$. Then
$$(\psi_2\circ\sigma)(y)=(\tau\circ\psi_1)(y)=\tau(x)$$
and thus ${\rm im}\,\tau\subset\psi_2({\rm im}\,\sigma)$.
Arguments like this are commonly known as proofs by diagram chase.
As $y\in\ker\rho$ it is correct to say $\psi_1(y)\in\psi_1(\ker\rho)$ as this is nothing else than applying $\psi_1$.
EDIT: Concerning the question below: consider
where the arrows are the respective inclusions and projections from the direct sums (and the identity for the rightmost column).$\tau$, in your notation, is surjective but $\sigma$ clearly not.
