Q1: Find $s_{12}$ if the nominal interest rate payable monthly is $5%$ per annum.
What I have done is: $$i^{(12)}=0.05$$ $$1+i=(1+i^{(12)}/12)^{12}$$ which leads to $$i=0.0512$$ $$s_{12}=((1+i)^{12}-1)/i=16.03$$
I wonder whether I am right to find $i$. Or should I use $i[12]$ which equals to $i^{(12)}/12$ instead.
Also,
Q2: A loan of £1000 is to be repaid by a level annuity, payable monthly in arrears for two years and calculated on the basis of an interest rate of $9%$ per annum. Calculate the monthly repayment.
In this question which $i$ should I use? $i$ or $i[12]$? I am quite confused.
I'm not exactly sure what your notation means so I'll just answer the question (Q2) and you can hopefully align the notations. Firstly we are asked for two years, this is $24$ months, the annual interest rate is $R=0.09$ which means that the monthly interest is $r=(1+R)^{(1/12)}-1 \approx 0.0072$. Now we will work with entirely monthly quantities.
Let $B(n)$ be the outstanding balance for month $n$, we have the boundary conditions $B(0)=1000$ and $B(23)=0$. For each month we make a payment of $C$, this covers the interest of $r\,B(n)$ and any excess goes to reducing the balance, thus we have $$B(n+1) = B(n) - (C-r\,B(n)) = B(n)(1+r) - C$$ This implies that $$\begin{align}B(n+2) &= B(n+1)(1+r) - C\\ &= (B(n)(1+r) - C))(1+r) - C\\ &= B(n)(1+r)^2 - C((1+r)-1)\end{align}$$ You should be able to see from this that in general we have $$\begin{align}B(n) &= B(0)(1+r)^n - C \sum_{j=0}^{n-1} (1+r)^j\\ &=B(0)(1+r)^n-C\frac{(1+r)^n-1}{r} \end{align}$$ We can substitute in our two boundary conditions to give $$0=B_0(1+r)^N - C\frac{(1+r)^N-1}{r}$$ $$\Rightarrow C = \frac{B_0\,r\,(1+r)^n}{(1+r^N)-1}$$ Which given the fact for us $B_0 = 1000, N=24$ and $r\approx 0.0072$ gives a payment rate of $$ C = 45.5 (3 \mathrm{s.f.}).$$