(a)

It helps to use #rightleftharpoons# instead:

#Zn_((s))+2Ag_((aq))^+rightleftharpoonsZn_((aq))^(2+)+2Ag_((s))#

Just by using Le Chatelier's Principle you can predict that by increasing #[Zn^(2+)]# the position of equilibrium will shift to the left thus decreasing #DeltaE_("cell")#.

(c)

#n# is the number of moles of electrons transferred which you can see must be 2 since we have :

#ZnrarrZn^(2+)+2e#

I'll attempt a fuller explanation:

The equation in the question refers to the electrode potential of the the 1/2 cell #E_("red")#, not #E_"cell"#. It should read:

#E_("cell")=E_("cell")^@-(RT)/(nF)lnQ#

At 298K this simplifies to:

#E_("cell")=E_("cell")^@-(0.05916)/(2)logQ#

#Q# is the reaction quotient and is given by:

#Q=([Zn^(2+)])/([Ag^+]^2)#

We can work out #E_("cell")^@# from standard electrode potentials:

#Zn^(2+)"/"Zn " "E^@=-0.76V#

#Ag^(+)"/"Ag" "E^@=+0.8V#

To find #E_("cell")^@# subtract the least +ve #E^@# from the most +ve #rArr#

#E_("cell")^@=0.8-(-0.76)=+1.56V#

If we had standard conditions i.e unit concentrations etc then #Q=1# so #"log"Q=0# and #E=E_("cell")-0=+1.56V#.

What happens if we increase #[Zn^(2+)]# to 2M as might happen in the question?

#E_("cell")=E_("cell")^@-(0.05916)/(2)log([2]/[1])#

#:.E_("cell")=+1.56-0.0089=+1.551V#

So you can see that there is a slight decrease as predicted.

I note the equation has now been corrected. The same reasoning applies. If #[Zn^(2+)]# is increased you can see that #E^@# is made less -ve so #E_("cell")# is reduced.