Let us start with two facts and a remark.
Fact 1. Let $(X,\mathcal{T})$ be a topological space and let $f \colon (X,\mathcal{T}) \to
\left[{-}\infty,{+}\infty\right]$. Then $f$ is lower semicontinuous if and only if, for every $\xi \in \mathbb{R}$, the lower level set $f^{-1}(\left[{-}\infty,\xi\right])$ is closed. Here, by lower semicontinuity, I mean: for every $x \in X$ and for every $\xi \in \left]-\infty,f(x)\right[$, there exists a neighborhood $V$ of $x$ in such that $(\forall y \in V)\; f(y) > \xi$.
Remark Lower semicontinuity goes with the topology on the domain of $f$. In particular, in your question, lower semicontinuous means "$f$ is lower semicontunuous wrt to the strong topology" whereas "weakly lower semicontinuous" means "$f$ is lower semicontinuous wrt to the weak topology on $H$." So I guess there is no way to avoid weak topology in the proof as it directly relates to the topologies on the domain.
Fact 2. Let $C$ be a convex subset of $H$ (in your question). Then $C$ is closed in the topology induced by the hilbertian norm of $H$ if and only if $C$ is closed in the weak topology.
Returning to your question and assume that $f$ is lower semicontinuous w.r.t the strong topology (induced by the norm of $H$) and that $f$ is convex. We must show that $f$ is weakly lower semicontinuous, i.e., $f$ is continuous when $H$ is equipped with the weak topology. Let us use Fact 1 to do this, i.e., take $\xi \in \mathbb{R}$ and show that $f^{-1}(\left[{-}\infty,\xi\right])$ is weakly closed. Since $f$ is convex, the set $f^{-1}(\left[{-}\infty,\xi\right])$ is convex. On the other hand, since $f$ is lsc w.r.t to the strong topology, the set $f^{-1}(\left[{-}\infty,\xi\right])$ is closed in the strong topology by Fact 1. Altogether, Fact 2 implies that it is indeed weakly closed.
So, we have shown that, for every $\xi \in \mathbb{R}$, the set $f^{-1}(\left[{-}\infty,\xi\right])$ is closed in the weak topology. In view of Fact 1, we conclude that $f$ is weakly lsc, i.e., lower semicontinuous when $H$ is equipped with the weak topology.
Best Answer
b) My Google responses on "weakly sequentially lower semicontinuous" query yield something about convexity. For instance, if $X$ is a normed space, and $f:X\to\overline{\mathbb R}$ is quasi-convex and (strongly) lower semicontinuous then $f$ is also weakly sequentially lower semicontinuous [1, Ex.3]. You can try to google for more such facts.
[1] Peter Holthe Hansen, 01716 Advanced Topics in Applied Functional Analysis, Assignment 10.
a) It seems the following.
An example of a continuous (and, hence, a lower semicontinuous) function $f:\ell_2\to\mathbb R$ such that $f$ is not weakly sequentially lower semicontinuous. For each point $x_0\in\ell_2$ define a function $f_{x_0}:\ell_2\to\mathbb R$ by putting $f_{x_0}(x)=\min\{\|x-x_0\|-1/3,0\}$ for each $x\in\ell_2$. For each natural number $n$ let $e_n$ be the standard $n$-th ort of the space $\ell_2$. Put $f=\sum_{n=1}^\infty f_{e_n}$. Then $f$ is continuous as a sum of a family of continuous functions with the locally finite family of the supports. From the other side, $f$ is not weakly sequentially lower semicontinuous, because $f(0)=0$, but the sequence $\{e_n\}$ weakly converges to $0$ and $f(e_n)=-1/3$ for each $n$.
An example of a weakly sequentially lower semicontinuous function $f:\ell_2\to\mathbb R$ such that $f$ is not weakly lower semicontinuous.
I found such a set $A=\{x_{nm}\}$ in Bill Johnson’s answer to Question “A point in the weak closure but not in the weak sequential closure”. The zero is the unique not weakly isolated point of the set $A$. So we may define a function $f:\ell_2\to\mathbb R$ by putting $f(0)=-1$, $f|A\setminus\{0\}=-2$ and $f|\ell_2\setminus A=0$.