chi squared additive property + subtraction property question so I got questions a + b right but when I got to c I made a couple of mistakes. Can someone tell me how to do c? I thought I was doing it right but I got the degree of freedom wrong and the symbols.

old answer: 

 A: The sum $U = \sum_{i=1}^5(X_i - \bar X)^2,$ where $\bar X$ is the mean of the
five observations, has $U \sim Chisq(\nu = 4).$ You say you have shown that.
Also, $X_6^2 \sim Chisq(\nu = 1).$  Then $T = U + X_6^2 \sim Chisq(\nu = 5),$
which can easily be shown using moment generating functions.
For a visual demonstration, the following R program generates a million
realizations of $U$ and $T.$ One indication that they have chi-squared
distributions with degrees of freedom 4 and 5, respectively, is that
$E(U) = 4$ and $E(T) = 5$ are well approximated by the simulations.
[Note: Perhaps surprisingly, the sample mean $A = \bar X$ and the sample variance
$S^2 = U/(n-1)$ are stocastically independent (for normal data only),
even though they are not functionally independent ($\bar X$ appears in
the formula for $U$). This is indicated
in the simulation because $Cor(A, U) = 0$ is well approximated.]
m = 10^6;  n = 5;  x = rnorm(m*n)
DTA = matrix(x, nrow=m)  # each row a sample of five
u = (n-1)*apply(DTA, 1, var) # numerators of m sample variances
a = rowMeans(DTA)  # vector of m sample means
v = rnorm(m)^2;  t = u + v
mean(u);  mean(t)
## 3.998416  # aprx E(U) = 4
## 4.999412  # aprx E(T) = 5
cor(a, v)
## -0.0003179211  #aprx C(A, U) = 0

Another verification is that the density curves of $Chisq(4)$ and
$Chisq(5)$ fit the histograms of the respective simulated distributions.

Here is the R code for the figure, in case it is of any interest to you.
par(mfrow=c(1,2))  # 2 panels per plot
 hist(u, prob=T, col="wheat", ylim=c(0,.2), main="U ~ CHISQ(4)")
    curve(dchisq(x,4), lwd=2, col="blue", add=T)
 hist(t, prob=T, col="wheat", main="T ~ CHISQ(5)")
    curve(dchisq(x,5), lwd=2, col="blue", add=T)
par(mfrow=c(1,1))  # return to default plot mode

Caution: Finally, you refer to 'subtraction' in your title. Just so there will be
no confusion: It is true that if $X \sim Chisq(m)$ and independently
$Y \sim Chisq(n),$ then $X + Y \sim Chisq(m+n),$ for positive integers $m$ and $n.$ However, it is not
true that $X - Y \sim Chisq(m - n),$ even if $m > n.$ 
