You have the linear equation (all log base 10)
log(y) = log(p) + t*log(q) = 4.8 + t*0.05
So
log(p) = 4.8
log(q) = 0.05
so
p = 10^4.8
q = 10^0.05 (must be slightly greater than 1, have you done 10^0.5?)
as its a "compound interest" type question (q-1)*100 would give the percentage "interest" each year. Each year, the price is multipled by q, so if it equals 1 it does not change. If its <1 the price decreases, if its >1 the price increases ... Similarly, taking logs, the log(v) graph would go up if log(q)>0 and down if log(q)<0 and be flat if log(q)=0.
Note for c) if you were unsure, you could get the 2010 price using the log(V) equation, then do 10^*. If you made an error on p and q, that part would not be affected.