Consider the following (mod 10)
$$0^2=0,1^2=1,2^2=4,3^2=9,4^2=6,5^2=5,6^2=6,7^2=9,8^2=4,9^2=1$$
Since the tens digit obviously doesn't matter for what the one digit is, these are all the ways that a square can end.
For the question about an even tens digit, just square $(\sum a_i 10^i)^2=a_0^2+2a_0a_1\cdot 10+\cdots$ You have an even number in front of the $10$, which switches to odd when $a_0^2$ has an odd tens digit. For the question about ending in a $5$ causing the $10$'s digit to be a $2$, notice that if $a_0=5$, then the first term is $25$ and the second is $100a_1$ which doesn't effect the tens spot.
You need to take the sum of the first $n$ odd squares, e.g.if $n=3$, then you need to add $1+9+25=35$. And, that does indeed equal $\frac{4}{3}n^3-\frac{1}{3}n$ for $n=3$: $\frac{4}{3}3^3-\frac{1}{3}3=\frac{4}{3}27-1=36-1=35$
Now, to prove that this is true in general using induction:
Base: $n=0$: $\frac{4}{3}0^3-\frac{1}{3}0=0-0=0$ which is indeed the sum of the first $0$ odd squares. Check!
Step: Assume that for some $k$ the sum of the first $k$ odd squares is $\frac{4}{3}k^3-\frac{1}{3}k$. Now let's consider the sum of the first $k+1$ odd squares, which is of course the sum of the first $k$ odd squares plus the $k+1$-th odd square, which is $(2k+1)^2$. So, by the inductive hypothesis the sum is $\frac{4}{3}k^3-\frac{1}{3}k+(2k+1)^2$, and now you need to verify that this does indeed equal $\frac{4}{3}(k+1)^3-\frac{1}{3}(k+1)$. Let's see:
$$\frac{4}{3}k^3-\frac{1}{3}k+(2k+1)^2=$$
$$\frac{4k^3-k+3(4k^2+4k+1)}{3}=$$
$$\frac{4k^3+12k^2+12k-k+3}{3}=$$
$$\frac{4k^3+12k^2+12k+4-k+3-4}{3}=$$
$$\frac{4(k^3+3k^2+3k+1)-(k+1)}{3}=$$
$$\frac{4(k+1)^3}{3}-\frac{1}{3}(k+1)$$
Best Answer
Since you're new to proofs, I'll sketch out the main idea of the proof and let you fill in the details. A good first step is to write down some variables, and state clearly what your claim is:
You want to prove that any consecutive perfect squares have odd difference; let $n^2$ be the first one, so that $(n + 1)^2$ is the larger one (make sure you can convince yourself that these really do represent consecutive squares). Now compute
$$(n + 1)^2 - n^2$$
and see what you conclude about it.