1.

power.t<-function(m){
              count.t<-0
              for (i in 1:100){
                x <- rnorm(25,5)
                y <- rnorm(25,5+m)
                if (t.test(x,y, alternative="l", var.equal=T)$p.value <.05) 
                    {count.t<-count.t+1}
                              }
              return(count.t/100)
                       } 

power.w<-function(m){
              count.w<-0
              for (i in 1:100){
                x <- rnorm(25,5)
                y <- rnorm(25,5+m)
                if (wilcox.test(x,y,alternative="l")$p.value <.05) 
                    {count.w<-count.w+1}    
                              }
              return(count.w/100)
                       } 


m<-seq(0,2,.25)
d<-length(m)
curve.t<-rep(0,d)
curve.w<-rep(0,d)
for (i in 1:d){
  curve.t[i]<-power.t(m[i])
  curve.w[i]<-power.w(m[i])
              }

plot(m, curve.t, type="l", lty=1)
points(m, curve.w, type="l", lty=2)
legend(1.5,0.2,c("t-test","Wilcoxon"),lty=c(1,2))


2. 

The 20 animals were randomized to 2 treatments (10 per treatment), so we have 
two independent samples. Alternative is two-sided.

(a) Assuming both underlying distributions are symmetric (which seems rather 
    dubious here), you could use a 2-sample t-test to assess the differences 
    between the population medians. Compare 2 SDs first. They are close. Then 
    do the "standard" 2-sample t-test (assuming common population variances).
    It yields a two-sided p-value of 0.006.

(b) Use of wilcox.test leads to a p-value of 0.02. 

(c) The sample size is not very large. So, t-test would do better if the 
    distributions are normal. However, there is not enough evidence to claim 
    normality (look at 2 qqplots for 2 treatments). So, probably prefer the 
    Wilcoxon in this situation.