= Solution
It suffices first to treat <unital algebras> and a <unital> <homomorphism>. A surjective <homomorphism> between <unital algebras> automatically preserves the identity. If $B=0$, <continuity> is immediate. In the nonunital case, extend to the forced <unitizations> by $T^+(a+\lambda1)=T(a)+\lambda1$. This is still surjective and the <unitization> of a semisimple algebra is semisimple: its radical projects to zero in its scalar quotient, and its remaining intersection with the original algebra is its original radical. <Continuity> of the extension implies <continuity> of the original map.
An <algebra homomorphism> decreases the <spectral radius>, even without any <continuity> assumption, because it takes an inverse of $\lambda1-a$ to an inverse of $\lambda1-T(a)$. Suppose $a_n\to0$ in $A$ and $T(a_n)\to b$ in $B$. Fix any $c\in B$ and choose $u,v\in A$ with $T(u)=b$, $T(v)=c$. Put
$$
w=vu,\quad w_n=va_n,\quad d=T(w)=cb,\quad d_n=T(w_n)\longrightarrow d.
$$
Consider the <polynomial> $q_n(z)=T((1-z)w_n+zw)=(1-z)d_n+zd$. Its value at one is exactly $d$. On the outer circle $|z|=R$,
$$
r_B(q_n(z))\le\|q_n(z)\|\le\|d\|+(1+R)\|d_n-d\|.
$$
On the inner circle $|z|=1/R$, spectral-radius decrease gives
$$
r_B(q_n(z))\le r_A((1-z)w_n+zw)
\le(1+1/R)\|w_n\|+\|w\|/R.
$$
Applying part (i) in $B$ and then letting $n\to\infty$, with $R$ fixed, gives
$$
r_B(cb)^2\le\frac{\|cb\|\|vu\|}{R}.
$$
This holds for every $R>1$, so \b[$r_B(cb)=0$ for every $c\in B$].
This forces $b$ into the <Jacobson radical>. To see the criterion directly, if $b$ were outside a <maximal left ideal> $L$, then $L+Bb=B$, so $1=l+cb$ for some $l\in L$, $c\in B$. The equality $r_B(cb)=0$ makes $1-cb$ invertible, but this element equals $l\in L$, impossible in a proper <left ideal>. Hence $b$ lies in every <maximal left ideal>. Since $B$ is semisimple, $b=0$.
We have shown that the separating space of $T$ is zero. If $a_n\to a$ and $T(a_n)\to y$, apply the result to $a_n-a$ to obtain $y=T(a)$. Its graph is <closed>, and the <closed graph theorem> now proves
$$
\boxed{T\text{ is continuous}.}
$$
The crucial estimate uses part (i) with an exactly fixed central value; it does not assume the false general assertion that the <spectral radius> is <continuous> under <norm> limits.
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