I don't know how to solve this exercise:
Compute the form of all implicit Runge Kutta method of order $3$ with 2 interrmediate steps from the integration formula $$\int_0^1\phi(t)\mathrm{d}t=\frac{a^2}{1+a^2}\phi\left(\frac{1}{2}+\frac{1}{2a\sqrt{3}}\right)+\frac{1}{1+a^2}\phi\left(\frac{1}{2}-\frac{a}{2\sqrt{3}}\right)$$
Now I know that we get RK from $$y(t_{n+1})=y(t_n)+\int_{t_n}^{t_{n+1}}f(t,y(t))dt\approx y(t_n)+h_n\sum\limits_{j=1}^q a_{ij}f(t_{nj},y(t_{n,j}))$$ But I fail to see how we apply this to the interval $[0,1]$ or what we have to do. I also don't know what semi-implicit means.
The order 3 conditions for a 2 stage Runge-Kutta method are, besides $c_j=a_{j1}+a_{j2}$, $$ b_1+b_2=1,\;b_1c_1+b_2c_2=\frac12,\;b_1c_1^2+b_2c_2^2=\frac13, \\ b_1a_{11}c_1+b_1a_{12}c_2+b_2a_{21}c_1+b_2a_{22}c_2=\frac16 $$ Now test these conditions on $$ c_1=\frac12-\frac1{2\sqrt3a},\; c_2=\frac12+\frac{a}{2\sqrt3},\; b_1=\frac{a^2}{1+a^2},\; b_2=\frac1{1+a^2}. $$ The first three conditions are easily seen to be satisfied. If semi-implicit means $a_{12}=0$ then $c_1=a_{11}$ and the last condition reduces to (using $a_{22}=c_2-a_{21}$) $$ \frac13+b_2a_{21}(c_1-c_2)=\frac16 $$ which uniquely determines $$ a_{21}=\frac1{6b_2(c_2-c_1)}=\frac{a}{2\sqrt3}. $$