I'm sorry, but I don't know how to do the second part of item 1 of this exercise:
Show that $\phi$ is l.s.c, that is, show that for each $\lambda \in \mathbb{R}$ set $$ [\phi \leq \lambda ] = \lbrace u \in H = L^{2}(0,1); \phi(u) \leq \lambda \rbrace $$ is closed.
Notice that $\varphi$ is LSC if it is LSC at every point and that every function is LSC outside its effective domain (where it is real valued). So you are left to show that $\varphi$ is LSC at each $u \in H^1(0,1)$, I suggest to use the sequential criterion, which is a characterization since the space is metric. Take such a $u$ and $u_j \longrightarrow u$ in $L^2(0,1)$, the goal is $$ \liminf_j \varphi(u_j) \geq \varphi(u). $$ Observe that for $v \in H^1(0,1)$, $$ \varphi(v) = \frac 1 2 ||v'||_{L^2(0,1)}^2. $$ If $\liminf_j \varphi(u_j) = \infty$ there is nothing to show so suppose without loss of generality it is a real number. You know that for a certain $\sigma : \mathbb N \longrightarrow \mathbb N$ increasing there holds $$ \liminf_j \varphi(u_j) = \lim_j \varphi(u_{\sigma(j)}) $$ so anytime we work "up to a subsequence" it does not matter. For this reason I suppose $(u_j)$ achieves the above limit. Now remember we took the liminf to be finite so for large $j$ the right hand side is finite so up to a subsequence $u_j \in H^1(0,1)$. Now the sequence $(||u_j'||_{L^2(0,1)})$ converges so is bounded and we have $(u_j')$ bounded in $L^2(0,1)$ that is a reflexive space (Hilbert) so up to another subsequence, $(u_j')$ converges for the weak topology to some $v \in L^2(0,1)$. And now we are happy because we know that in a normed space the norm is weakly LSC (because convex and strongly LSC) so that $$ ||v||_{L^2(0,1)} \leq \liminf_j ||u_j'||_{L^2(0,1)}. $$ Now to conclude show that $v = u'$, realize that $(u_j')$ goes to $u'$ weakly in $L^2(0,1)$ and apply a continuous non-decreasing function to the last inequality.